In calculus, limits are used to define both the derivative and the integral. They also help us understand the behavior of functions as they approach specific points or as they go to infinity. Continuity ensures that a function behaves smoothly and predictably at every point in its domain.
1. Definition of a Limit
The limit of a function $f(x)$ as $x$ approaches some value $a$ is the value that $f(x)$ gets closer to as $x$ gets closer to $a$. This is written as:
$ \lim_{x \to a} f(x) = L $
This means that as $x$ gets closer to $a$ (from either the left or right), $f(x)$ approaches the value $L$.
Example 1: Finding a Limit
Problem:
Find $ \lim_{x \to 3} (2x + 1) $.
Answer:
Step 1: Given Data:
$ f(x) = 2x + 1 $
Step 2: Solution:
Substitute $x = 3$ into the function:
$ f(3) = 2(3) + 1 $
$ = 6 + 1 $
$ = 7 $
Step 3: Final Answer:
$ \lim_{x \to 3} (2x + 1) = 7 $
2. One-Sided Limits
A one-sided limit considers the behavior of a function as $x$ approaches $a$ from only one side, either from the left or the right.
- Left-hand limit: $ \lim_{x \to a^{-}} f(x) $
This is the limit of $f(x)$ as $x$ approaches $a$ from the left side. - Right-hand limit: $ \lim_{x \to a^{+}} f(x) $
This is the limit of $f(x)$ as $x$ approaches $a$ from the right side.
Example 2: One-Sided Limit
Problem:
Find $ \lim_{x \to 2^{-}} (3x – 4) $ and $ \lim_{x \to 2^{+}} (3x – 4) $.
Answer:
Step 1: Given Data:
$ f(x) = 3x – 4 $
Step 2: Solution (left-hand limit):
Substitute $x = 2$ from the left:
$ \lim_{x \to 2^{-}} (3x – 4) = 3(2) – 4 $
$ = 6 – 4 $
$ = 2 $
Step 3: Final Answer (left-hand limit):
$ \lim_{x \to 2^{-}} (3x – 4) = 2 $
Step 4: Solution (right-hand limit):
Substitute $x = 2$ from the right:
$ \lim_{x \to 2^{+}} (3x – 4) = 3(2) – 4 $
$ = 6 – 4 $
$ = 2 $
Step 5: Final Answer (right-hand limit):
$ \lim_{x \to 2^{+}} (3x – 4) = 2 $
3. Continuity
A function is said to be continuous at a point $x = a$ if the following three conditions are met:
- $ f(a) $ is defined.
- $ \lim_{x \to a} f(x) $ exists.
- $ \lim_{x \to a} f(x) = f(a) $.
If these three conditions hold, we say that $f(x)$ is continuous at $x = a$.
Example 3: Checking Continuity
Problem:
Determine if the function $f(x) = 2x + 3$ is continuous at $x = 1$.
Answer:
Step 1: Given Data:
$ f(x) = 2x + 3 $
Step 2: Solution (checking conditions for continuity):
- First, check if $f(1)$ is defined:
$ f(1) = 2(1) + 3 $
$ = 2 + 3 $
$ = 5 $
Since $f(1)$ is defined, condition 1 is satisfied. - Next, find $ \lim_{x \to 1} f(x) $:
$ \lim_{x \to 1} (2x + 3) = 2(1) + 3 $
$ = 5 $
Since the limit exists, condition 2 is satisfied. - Finally, check if the limit equals the value of the function:
$ \lim_{x \to 1} f(x) = f(1) = 5 $
Step 3: Final Answer:
Since all three conditions are met, $f(x)$ is continuous at $x = 1$.
4. Limits at Infinity
A limit at infinity describes the behavior of a function as $x$ approaches infinity ($\infty$) or negative infinity ($-\infty$). For example, if the function $f(x)$ approaches some value $L$ as $x$ increases without bound, we write:
$ \lim_{x \to \infty} f(x) = L $
Similarly, if $x$ approaches negative infinity:
$ \lim_{x \to -\infty} f(x) = L $
Example 4: Limit at Infinity
Problem:
Find $ \lim_{x \to \infty} \frac{3}{x} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{3}{x} $
Step 2: Solution:
As $x$ increases without bound, $\frac{3}{x}$ gets closer to zero.
$ \lim_{x \to \infty} \frac{3}{x} = 0 $
Step 3: Final Answer:
$ \lim_{x \to \infty} \frac{3}{x} = 0 $
5. Special Limits
There are several important limits that appear frequently in calculus. One of the most famous special limits is:
$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 $
This result is often used in conjunction with L’Hôpital’s rule or when working with trigonometric limits.
Example 5: Special Limit
Problem:
Find $ \lim_{x \to 0} \frac{\sin(2x)}{x} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{\sin(2x)}{x} $
Step 2: Solution:
We can rewrite the limit as:
$ \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{2 \sin(2x)}{2x} $
We know that $ \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 $, so we multiply by 2:
$ = 2 \times 1 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 $
Conclusion
Limits are fundamental in calculus, providing the foundation for defining derivatives and integrals. They allow us to analyze the behavior of functions as they approach specific values or infinity. Continuity, on the other hand, ensures that functions behave smoothly at every point in their domain. Understanding how to compute and interpret limits is essential for working with continuous functions and their applications.
Question And Answer Library
Example 1
Problem:
Find $ \lim_{x \to 2} (3x + 4) $.
Answer:
Step 1: Given Data:
$ f(x) = 3x + 4 $
Step 2: Solution:
Substitute $ x = 2 $ into the function:
$ f(2) = 3(2) + 4 = 6 + 4 = 10 $
Step 3: Final Answer:
$ \lim_{x \to 2} (3x + 4) = 10 $
Example 2
Problem:
Find $ \lim_{x \to -1} (x^2 – 3x + 2) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 – 3x + 2 $
Step 2: Solution:
Substitute $ x = -1 $ into the function:
$ f(-1) = (-1)^2 – 3(-1) + 2 = 1 + 3 + 2 = 6 $
Step 3: Final Answer:
$ \lim_{x \to -1} (x^2 – 3x + 2) = 6 $
Example 3
Problem:
Find $ \lim_{x \to 4} (x^2 – 16) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 – 16 $
Step 2: Solution:
Substitute $ x = 4 $ into the function:
$ f(4) = 4^2 – 16 = 16 – 16 = 0 $
Step 3: Final Answer:
$ \lim_{x \to 4} (x^2 – 16) = 0 $
Example 4
Problem:
Find $ \lim_{x \to 0} \frac{1}{x+1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{1}{x+1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{1}{0 + 1} = \frac{1}{1} = 1 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{1}{x+1} = 1 $
Example 5
Problem:
Find $ \lim_{x \to 1} (4x – x^2) $.
Answer:
Step 1: Given Data:
$ f(x) = 4x – x^2 $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = 4(1) – (1)^2 = 4 – 1 = 3 $
Step 3: Final Answer:
$ \lim_{x \to 1} (4x – x^2) = 3 $
Example 6
Problem:
Find $ \lim_{x \to 3} (x^2 + 2x – 5) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 + 2x – 5 $
Step 2: Solution:
Substitute $ x = 3 $ into the function:
$ f(3) = (3)^2 + 2(3) – 5 = 9 + 6 – 5 = 10 $
Step 3: Final Answer:
$ \lim_{x \to 3} (x^2 + 2x – 5) = 10 $
Example 7
Problem:
Find $ \lim_{x \to -2} (x^3 + 4) $.
Answer:
Step 1: Given Data:
$ f(x) = x^3 + 4 $
Step 2: Solution:
Substitute $ x = -2 $ into the function:
$ f(-2) = (-2)^3 + 4 = -8 + 4 = -4 $
Step 3: Final Answer:
$ \lim_{x \to -2} (x^3 + 4) = -4 $
Example 8
Problem:
Find $ \lim_{x \to 0} (x^2 + 2x + 1) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 + 2x + 1 $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = (0)^2 + 2(0) + 1 = 1 $
Step 3: Final Answer:
$ \lim_{x \to 0} (x^2 + 2x + 1) = 1 $
Example 9
Problem:
Find $ \lim_{x \to 5} (x + 7) $.
Answer:
Step 1: Given Data:
$ f(x) = x + 7 $
Step 2: Solution:
Substitute $ x = 5 $ into the function:
$ f(5) = 5 + 7 = 12 $
Step 3: Final Answer:
$ \lim_{x \to 5} (x + 7) = 12 $
Example 10
Problem:
Find $ \lim_{x \to 1} \frac{1}{x^2 – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{1}{x^2 – 1} $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = \frac{1}{(1)^2 – 1} = \frac{1}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to 1} \frac{1}{x^2 – 1} $ does not exist.
Example 11
Problem:
Find $ \lim_{x \to 2} (x^2 – 4x + 5) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 – 4x + 5 $
Step 2: Solution:
Substitute $ x = 2 $ into the function:
$ f(2) = (2)^2 – 4(2) + 5 = 4 – 8 + 5 = 1 $
Step 3: Final Answer:
$ \lim_{x \to 2} (x^2 – 4x + 5) = 1 $
Example 12
Problem:
Find $ \lim_{x \to 0} \frac{2x}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x}{x + 1} = 0 $
Example 13
Problem:
Find $ \lim_{x \to -3} (x^2 + 9) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 + 9 $
Step 2: Solution:
Substitute $ x = -3 $ into the function:
$ f(-3) = (-3)^2 + 9 = 9 + 9 = 18 $
Step 3: Final Answer:
$ \lim_{x \to -3} (x^2 + 9) = 18 $
Example 14
Problem:
Find $ \lim_{x \to 1} (5x – 2) $.
Answer:
Step 1: Given Data:
$ f(x) = 5x – 2 $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = 5(1) – 2 = 5 – 2 = 3 $
Step 3: Final Answer:
$ \lim_{x \to 1} (5x – 2) = 3 $
Example 15
Problem:
Find $ \lim_{x \to 4} (x^3 – 3x^2 + 5) $.
Answer:
Step 1: Given Data:
$ f(x) = x^3 – 3x^2 + 5 $
Step 2: Solution:
Substitute $ x = 4 $ into the function:
$ f(4) = (4)^3 – 3(4)^2 + 5 = 64 – 48 + 5 = 21 $
Step 3: Final Answer:
$ \lim_{x \to 4} (x^3 – 3x^2 + 5) = 21 $
Example 16
Problem:
Find $ \lim_{x \to -2} (x^4 – x^2 + 1) $.
Answer:
Step 1: Given Data:
$ f(x) = x^4 – x^2 + 1 $
Step 2: Solution:
Substitute $ x = -2 $ into the function:
$ f(-2) = (-2)^4 – (-2)^2 + 1 = 16 – 4 + 1 = 13 $
Step 3: Final Answer:
$ \lim_{x \to -2} (x^4 – x^2 + 1) = 13 $
Example 17
Problem:
Find $ \lim_{x \to 3} \frac{x^2 + 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 1}{x – 1} $
Step 2: Solution:
Substitute $ x = 3 $ into the function:
$ f(3) = \frac{(3)^2 + 1}{3 – 1} = \frac{9 + 1}{2} = \frac{10}{2} = 5 $
Step 3: Final Answer:
$ \lim_{x \to 3} \frac{x^2 + 1}{x – 1} = 5 $
Example 18
Problem:
Find $ \lim_{x \to 1} \frac{1}{x} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{1}{x} $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = \frac{1}{1} = 1 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{1}{x} = 1 $
Example 19
Problem:
Find $ \lim_{x \to -1} (x^3 + 2x) $.
Answer:
Step 1: Given Data:
$ f(x) = x^3 + 2x $
Step 2: Solution:
Substitute $ x = -1 $ into the function:
$ f(-1) = (-1)^3 + 2(-1) = -1 – 2 = -3 $
Step 3: Final Answer:
$ \lim_{x \to -1} (x^3 + 2x) = -3 $
Example 20
Problem:
Find $ \lim_{x \to 2} \frac{x^2 – 4}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 4}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x + 2)}{x – 2} $
Cancel the common factor:
$ x + 2 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = 2 + 2 = 4 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^2 – 4}{x – 2} = 4 $
Example 21
Problem:
Find $ \lim_{x \to 0} (5x^3 – x^2) $.
Answer:
Step 1: Given Data:
$ f(x) = 5x^3 – x^2 $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = 5(0)^3 – (0)^2 = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} (5x^3 – x^2) = 0 $
Example 22
Problem:
Find $ \lim_{x \to 1} (x^2 – x + 1) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 – x + 1 $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = (1)^2 – 1 + 1 = 1 $
Step 3: Final Answer:
$ \lim_{x \to 1} (x^2 – x + 1) = 1 $
Example 23
Problem:
Find $ \lim_{x \to 2} \frac{2x – 3}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x – 3}{x + 1} $
Step 2: Solution:
Substitute $ x = 2 $ into the function:
$ f(2) = \frac{2(2) – 3}{2 + 1} = \frac{4 – 3}{3} = \frac{1}{3} $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{2x – 3}{x + 1} = \frac{1}{3} $
Example 24
Problem:
Find $ \lim_{x \to 0} \frac{x}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x}{x + 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0}{0 + 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x}{x + 2} = 0 $
Example 25
Problem:
Find $ \lim_{x \to -1} \frac{x + 1}{x^2 – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x + 1}{x^2 – 1} $
Step 2: Solution:
Substitute $ x = -1 $ into the function:
$ f(-1) = \frac{-1 + 1}{(-1)^2 – 1} = \frac{0}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to -1} \frac{x + 1}{x^2 – 1} $ does not exist.
Example 26
Problem:
Find $ \lim_{x \to 0} (4x^2 – 3x) $.
Answer:
Step 1: Given Data:
$ f(x) = 4x^2 – 3x $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = 4(0)^2 – 3(0) = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} (4x^2 – 3x) = 0 $
Example 27
Problem:
Find $ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 1}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 1)(x + 1)}{x – 1} $
Cancel the common factor:
$ x + 1 $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1 + 1 = 2 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} = 2 $
Example 28
Problem:
Find $ \lim_{x \to 2} (5x^2 – 6x + 2) $.
Answer:
Step 1: Given Data:
$ f(x) = 5x^2 – 6x + 2 $
Step 2: Solution:
Substitute $ x = 2 $ into the function:
$ f(2) = 5(2)^2 – 6(2) + 2 = 20 – 12 + 2 = 10 $
Step 3: Final Answer:
$ \lim_{x \to 2} (5x^2 – 6x + 2) = 10 $
Example 29
Problem:
Find $ \lim_{x \to -2} (3x^2 – 4x + 1) $.
Answer:
Step 1: Given Data:
$ f(x) = 3x^2 – 4x + 1 $
Step 2: Solution:
Substitute $ x = -2 $ into the function:
$ f(-2) = 3(-2)^2 – 4(-2) + 1 = 12 + 8 + 1 = 21 $
Step 3: Final Answer:
$ \lim_{x \to -2} (3x^2 – 4x + 1) = 21 $
Example 30
Problem:
Find $ \lim_{x \to 0} \frac{2x}{x^2 + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x}{x^2 + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)}{0^2 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x}{x^2 + 1} = 0 $
Example 31
Problem:
Find $ \lim_{x \to 3} (2x^2 – 5x + 7) $.
Answer:
Step 1: Given Data:
$ f(x) = 2x^2 – 5x + 7 $
Step 2: Solution:
Substitute $ x = 3 $ into the function:
$ f(3) = 2(3)^2 – 5(3) + 7 = 18 – 15 + 7 = 10 $
Step 3: Final Answer:
$ \lim_{x \to 3} (2x^2 – 5x + 7) = 10 $
Example 32
Problem:
Find $ \lim_{x \to 0} \frac{x^2}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0^2}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^2}{x + 1} = 0 $
Example 33
Problem:
Find $ \lim_{x \to 2} \frac{x – 2}{x^2 – 4} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x – 2}{x^2 – 4} $
Step 2: Solution:
Factor the denominator:
$ \frac{x – 2}{(x – 2)(x + 2)} $
Cancel the common factor:
$ \frac{1}{x + 2} $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = \frac{1}{2 + 2} = \frac{1}{4} $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x – 2}{x^2 – 4} = \frac{1}{4} $
Example 34
Problem:
Find $ \lim_{x \to -1} \frac{2x + 3}{x^2 – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x + 3}{x^2 – 1} $
Step 2: Solution:
Substitute $ x = -1 $ into the function:
$ f(-1) = \frac{2(-1) + 3}{(-1)^2 – 1} = \frac{-2 + 3}{1 – 1} = \frac{1}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to -1} \frac{2x + 3}{x^2 – 1} $ does not exist.
Example 35
Problem:
Find $ \lim_{x \to 0} \frac{x}{x^3 + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x}{x^3 + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0}{0^3 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x}{x^3 + 1} = 0 $
Example 36
Problem:
Find $ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 1}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 1)(x + 1)}{x – 1} $
Cancel the common factor:
$ x + 1 $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1 + 1 = 2 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} = 2 $
Example 37
Problem:
Find $ \lim_{x \to 2} \frac{x^2 – 2x}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 2x}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{x(x – 2)}{x – 2} $
Cancel the common factor:
$ x $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = 2 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^2 – 2x}{x – 2} = 2 $
Example 38
Problem:
Find $ \lim_{x \to 0} \frac{3x}{x^2 + 4} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{3x}{x^2 + 4} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{3(0)}{0^2 + 4} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{3x}{x^2 + 4} = 0 $
Example 39
Problem:
Find $ \lim_{x \to 2} (x^2 – 4) $.
Answer:
Step 1: Given Data:
$ f(x) = x^2 – 4 $
Step 2: Solution:
Substitute $ x = 2 $ into the function:
$ f(2) = (2)^2 – 4 = 4 – 4 = 0 $
Step 3: Final Answer:
$ \lim_{x \to 2} (x^2 – 4) = 0 $
Example 40
Problem:
Find $ \lim_{x \to -1} (x^3 + 2x^2 – 1) $.
Answer:
Step 1: Given Data:
$ f(x) = x^3 + 2x^2 – 1 $
Step 2: Solution:
Substitute $ x = -1 $ into the function:
$ f(-1) = (-1)^3 + 2(-1)^2 – 1 = -1 + 2 – 1 = 0 $
Step 3: Final Answer:
$ \lim_{x \to -1} (x^3 + 2x^2 – 1) = 0 $
Example 41
Problem:
Find $ \lim_{x \to 3} \frac{2x^2 – 3x}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^2 – 3x}{x – 1} $
Step 2: Solution:
Substitute $ x = 3 $ into the function:
$ f(3) = \frac{2(3)^2 – 3(3)}{3 – 1} = \frac{18 – 9}{2} = \frac{9}{2} $
Step 3: Final Answer:
$ \lim_{x \to 3} \frac{2x^2 – 3x}{x – 1} = \frac{9}{2} $
Example 42
Problem:
Find $ \lim_{x \to 0} \frac{x^2}{2x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2}{2x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0^2}{2(0) + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^2}{2x + 1} = 0 $
Example 43
Problem:
Find $ \lim_{x \to 4} \frac{x^2 – 16}{x – 4} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 16}{x – 4} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 4)(x + 4)}{x – 4} $
Cancel the common factor:
$ x + 4 $
Substitute $ x = 4 $ into the simplified expression:
$ f(4) = 4 + 4 = 8 $
Step 3: Final Answer:
$ \lim_{x \to 4} \frac{x^2 – 16}{x – 4} = 8 $
Example 44
Problem:
Find $ \lim_{x \to 0} \frac{3x^2}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{3x^2}{x + 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{3(0)^2}{0 + 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{3x^2}{x + 2} = 0 $
Example 45
Problem:
Find $ \lim_{x \to -2} \frac{x^2 – 4}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 4}{x + 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x + 2)}{x + 2} $
Cancel the common factor:
$ x – 2 $
Substitute $ x = -2 $ into the simplified expression:
$ f(-2) = -2 – 2 = -4 $
Step 3: Final Answer:
$ \lim_{x \to -2} \frac{x^2 – 4}{x + 2} = -4 $
Example 46
Problem:
Find $ \lim_{x \to 1} \frac{x^2 – x}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – x}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{x(x – 1)}{x – 1} $
Cancel the common factor:
$ x $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^2 – x}{x – 1} = 1 $
Example 47
Problem:
Find $ \lim_{x \to 0} \frac{4x}{x^2 + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{4x}{x^2 + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{4(0)}{0^2 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{4x}{x^2 + 1} = 0 $
Example 48
Problem:
Find $ \lim_{x \to 2} \frac{x^2 + 3x – 4}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 3x – 4}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x + 4)}{x – 2} $
Cancel the common factor:
$ x + 4 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = 2 + 4 = 6 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^2 + 3x – 4}{x – 2} = 6 $
Example 49
Problem:
Find $ \lim_{x \to 1} \frac{5x^2 – 4x}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{5x^2 – 4x}{x – 1} $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = \frac{5(1)^2 – 4(1)}{1 – 1} = \frac{5 – 4}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to 1} \frac{5x^2 – 4x}{x – 1} $ does not exist.
Example 50
Problem:
Find $ \lim_{x \to 0} \frac{2x^3}{x^2 + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^3}{x^2 + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)^3}{0^2 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x^3}{x^2 + 1} = 0 $
Example 51
Problem:
Find $ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 9}{x – 3} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 3)(x + 3)}{x – 3} $
Cancel the common factor:
$ x + 3 $
Substitute $ x = 3 $ into the simplified expression:
$ f(3) = 3 + 3 = 6 $
Step 3: Final Answer:
$ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 6 $
Example 52
Problem:
Find $ \lim_{x \to 0} \frac{x^2 + 2x}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 2x}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0^2 + 2(0)}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^2 + 2x}{x + 1} = 0 $
Example 53
Problem:
Find $ \lim_{x \to 4} \frac{x – 4}{x^2 – 16} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x – 4}{x^2 – 16} $
Step 2: Solution:
Factor the denominator:
$ \frac{x – 4}{(x – 4)(x + 4)} $
Cancel the common factor:
$ \frac{1}{x + 4} $
Substitute $ x = 4 $ into the simplified expression:
$ f(4) = \frac{1}{4 + 4} = \frac{1}{8} $
Step 3: Final Answer:
$ \lim_{x \to 4} \frac{x – 4}{x^2 – 16} = \frac{1}{8} $
Example 54
Problem:
Find $ \lim_{x \to -3} \frac{2x^2 + 5x + 3}{x + 3} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^2 + 5x + 3}{x + 3} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 3)(2x + 1)}{x + 3} $
Cancel the common factor:
$ 2x + 1 $
Substitute $ x = -3 $ into the simplified expression:
$ f(-3) = 2(-3) + 1 = -6 + 1 = -5 $
Step 3: Final Answer:
$ \lim_{x \to -3} \frac{2x^2 + 5x + 3}{x + 3} = -5 $
Example 55
Problem:
Find $ \lim_{x \to 0} \frac{5x}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{5x}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{5(0)}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{5x}{x + 1} = 0 $
Example 56
Problem:
Find $ \lim_{x \to 1} \frac{x^3 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 1}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 1)(x^2 + x + 1)}{x – 1} $
Cancel the common factor:
$ x^2 + x + 1 $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1^2 + 1 + 1 = 3 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^3 – 1}{x – 1} = 3 $
Example 57
Problem:
Find $ \lim_{x \to 0} \frac{x^2}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2}{x – 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0^2}{0 – 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^2}{x – 2} = 0 $
Example 58
Problem:
Find $ \lim_{x \to 0} \frac{3x}{x + 5} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{3x}{x + 5} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{3(0)}{0 + 5} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{3x}{x + 5} = 0 $
Example 59
Problem:
Find $ \lim_{x \to 2} \frac{x^2 – 4}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 4}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x + 2)}{x – 2} $
Cancel the common factor:
$ x + 2 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = 2 + 2 = 4 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^2 – 4}{x – 2} = 4 $
Example 60
Problem:
Find $ \lim_{x \to 1} \frac{x^3 – x}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – x}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{x(x – 1)(x + 1)}{x – 1} $
Cancel the common factor:
$ x(x + 1) $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1(1 + 1) = 2 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^3 – x}{x – 1} = 2 $
Example 61
Problem:
Find $ \lim_{x \to 0} \frac{2x}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x}{x + 1} = 0 $
Example 62
Problem:
Find $ \lim_{x \to 4} \frac{x^2 – 16}{x^2 – 4x} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 16}{x^2 – 4x} $
Step 2: Solution:
Factor both the numerator and denominator:
$ \frac{(x – 4)(x + 4)}{x(x – 4)} $
Cancel the common factor:
$ \frac{x + 4}{x} $
Substitute $ x = 4 $ into the simplified expression:
$ f(4) = \frac{4 + 4}{4} = \frac{8}{4} = 2 $
Step 3: Final Answer:
$ \lim_{x \to 4} \frac{x^2 – 16}{x^2 – 4x} = 2 $
Example 63
Problem:
Find $ \lim_{x \to 0} \frac{x^3}{x^2 + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3}{x^2 + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0^3}{0^2 + 1} = \frac{0}{1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^3}{x^2 + 1} = 0 $
Example 64
Problem:
Find $ \lim_{x \to 1} \frac{2x^2 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^2 – 1}{x – 1} $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = \frac{2(1)^2 – 1}{1 – 1} = \frac{2 – 1}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to 1} \frac{2x^2 – 1}{x – 1} $ does not exist.
Example 65
Problem:
Find $ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 8}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x^2 + 2x + 4)}{x – 2} $
Cancel the common factor:
$ x^2 + 2x + 4 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = (2)^2 + 2(2) + 4 = 4 + 4 + 4 = 12 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} = 12 $
Example 66
Problem:
Find $ \lim_{x \to -1} \frac{x^2 + x}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + x}{x + 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{x(x + 1)}{x + 1} $
Cancel the common factor:
$ x $
Substitute $ x = -1 $ into the simplified expression:
$ f(-1) = -1 $
Step 3: Final Answer:
$ \lim_{x \to -1} \frac{x^2 + x}{x + 1} = -1 $
Example 67
Problem:
Find $ \lim_{x \to 0} \frac{x^3 + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 1}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{(0)^3 + 1}{0 + 1} = \frac{1}{1} = 1 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^3 + 1}{x + 1} = 1 $
Example 68
Problem:
Find $ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 9}{x – 3} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 3)(x + 3)}{x – 3} $
Cancel the common factor:
$ x + 3 $
Substitute $ x = 3 $ into the simplified expression:
$ f(3) = 3 + 3 = 6 $
Step 3: Final Answer:
$ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 6 $
Example 69
Problem:
Find $ \lim_{x \to -2} \frac{x^2 + 4}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 4}{x + 2} $
Step 2: Solution:
Substitute $ x = -2 $ into the function:
$ f(-2) = \frac{(-2)^2 + 4}{-2 + 2} = \frac{4 + 4}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to -2} \frac{x^2 + 4}{x + 2} $ does not exist.
Example 70
Problem:
Find $ \lim_{x \to 1} \frac{x – 1}{x^2 – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x – 1}{x^2 – 1} $
Step 2: Solution:
Factor the denominator:
$ \frac{x – 1}{(x – 1)(x + 1)} $
Cancel the common factor:
$ \frac{1}{x + 1} $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = \frac{1}{1 + 1} = \frac{1}{2} $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x – 1}{x^2 – 1} = \frac{1}{2} $
Example 71
Problem:
Find $ \lim_{x \to 0} \frac{x^2 + 2x + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 2x + 1}{x + 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 1)(x + 1)}{x + 1} $
Cancel the common factor:
$ x + 1 $
Substitute $ x = 0 $ into the simplified expression:
$ f(0) = 0 + 1 = 1 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^2 + 2x + 1}{x + 1} = 1 $
Example 72
Problem:
Find $ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 1}{x + 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 1)(x^2 – x + 1)}{x + 1} $
Cancel the common factor:
$ x^2 – x + 1 $
Substitute $ x = -1 $ into the simplified expression:
$ f(-1) = (-1)^2 – (-1) + 1 = 1 + 1 + 1 = 3 $
Step 3: Final Answer:
$ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} = 3 $
Example 73
Problem:
Find $ \lim_{x \to 0} \frac{2x^2}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^2}{x + 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)^2}{0 + 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x^2}{x + 2} = 0 $
Example 74
Problem:
Find $ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 8}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x^2 + 2x + 4)}{x – 2} $
Cancel the common factor:
$ x^2 + 2x + 4 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = (2)^2 + 2(2) + 4 = 4 + 4 + 4 = 12 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} = 12 $
Example 75
Problem:
Find $ \lim_{x \to 0} \frac{4x^2 + 2x}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{4x^2 + 2x}{x + 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{4(0)^2 + 2(0)}{0 + 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{4x^2 + 2x}{x + 2} = 0 $
Example 76
Problem:
Find $ \lim_{x \to -2} \frac{x^2 + 4x}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 4x}{x + 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{x(x + 4)}{x + 2} $
Substitute $ x = -2 $ into the simplified expression:
$ f(-2) = \frac{(-2)^2 + 4(-2)}{-2 + 2} = \frac{4 – 8}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to -2} \frac{x^2 + 4x}{x + 2} $ does not exist.
Example 77
Problem:
Find $ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 1}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 1)(x + 1)}{x – 1} $
Cancel the common factor:
$ x + 1 $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1 + 1 = 2 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} = 2 $
Example 78
Problem:
Find $ \lim_{x \to 0} \frac{5x^3 + 2x}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{5x^3 + 2x}{x + 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{5(0)^3 + 2(0)}{0 + 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{5x^3 + 2x}{x + 2} = 0 $
Example 79
Problem:
Find $ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 9}{x – 3} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 3)(x + 3)}{x – 3} $
Cancel the common factor:
$ x + 3 $
Substitute $ x = 3 $ into the simplified expression:
$ f(3) = 3 + 3 = 6 $
Step 3: Final Answer:
$ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 6 $
Example 80
Problem:
Find $ \lim_{x \to -2} \frac{x^3 + 8}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 8}{x + 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 2)(x^2 – 2x + 4)}{x + 2} $
Cancel the common factor:
$ x^2 – 2x + 4 $
Substitute $ x = -2 $ into the simplified expression:
$ f(-2) = (-2)^2 – 2(-2) + 4 = 4 + 4 + 4 = 12 $
Step 3: Final Answer:
$ \lim_{x \to -2} \frac{x^3 + 8}{x + 2} = 12 $
Example 81
Problem:
Find $ \lim_{x \to 1} \frac{x^2 – 1}{x^2 – 2x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 1}{x^2 – 2x + 1} $
Step 2: Solution:
Factor both the numerator and denominator:
$ \frac{(x – 1)(x + 1)}{(x – 1)^2} $
Cancel the common factor:
$ \frac{x + 1}{x – 1} $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = \frac{1 + 1}{1 – 1} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to 1} \frac{x^2 – 1}{x^2 – 2x + 1} $ does not exist.
Example 82
Problem:
Find $ \lim_{x \to 0} \frac{2x^2 + 3x}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^2 + 3x}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)^2 + 3(0)}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x^2 + 3x}{x + 1} = 0 $
Example 83
Problem:
Find $ \lim_{x \to -1} \frac{x^2 + 2x + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 2x + 1}{x + 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 1)(x + 1)}{x + 1} $
Cancel the common factor:
$ x + 1 $
Substitute $ x = -1 $ into the simplified expression:
$ f(-1) = -1 + 1 = 0 $
Step 3: Final Answer:
$ \lim_{x \to -1} \frac{x^2 + 2x + 1}{x + 1} = 0 $
Example 84
Problem:
Find $ \lim_{x \to 2} \frac{x^3 – 8}{x^2 – 4} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 8}{x^2 – 4} $
Step 2: Solution:
Factor both the numerator and denominator:
$ \frac{(x – 2)(x^2 + 2x + 4)}{(x – 2)(x + 2)} $
Cancel the common factor:
$ \frac{x^2 + 2x + 4}{x + 2} $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = \frac{(2)^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = 3 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^3 – 8}{x^2 – 4} = 3 $
Example 85
Problem:
Find $ \lim_{x \to -2} \frac{x^3 + 8}{x^2 – 4} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 8}{x^2 – 4} $
Step 2: Solution:
Factor both the numerator and denominator:
$ \frac{(x + 2)(x^2 – 2x + 4)}{(x + 2)(x – 2)} $
Cancel the common factor:
$ \frac{x^2 – 2x + 4}{x – 2} $
Substitute $ x = -2 $ into the simplified expression:
$ f(-2) = \frac{(-2)^2 – 2(-2) + 4}{-2 – 2} = \frac{4 + 4 + 4}{-4} = -3 $
Step 3: Final Answer:
$ \lim_{x \to -2} \frac{x^3 + 8}{x^2 – 4} = -3 $
Example 86
Problem:
Find $ \lim_{x \to 0} \frac{4x^2 + 5x}{x^2 + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{4x^2 + 5x}{x^2 + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{4(0)^2 + 5(0)}{0^2 + 1} = \frac{0}{1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{4x^2 + 5x}{x^2 + 1} = 0 $
Example 87
Problem:
Find $ \lim_{x \to 1} \frac{x^3 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 1}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 1)(x^2 + x + 1)}{x – 1} $
Cancel the common factor:
$ x^2 + x + 1 $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = (1)^2 + 1 + 1 = 3 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^3 – 1}{x – 1} = 3 $
Example 88
Problem:
Find $ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 8}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x^2 + 2x + 4)}{x – 2} $
Cancel the common factor:
$ x^2 + 2x + 4 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = 2^2 + 2(2) + 4 = 12 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} = 12 $
Example 89
Problem:
Find $ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 9}{x – 3} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 3)(x + 3)}{x – 3} $
Cancel the common factor:
$ x + 3 $
Substitute $ x = 3 $ into the simplified expression:
$ f(3) = 3 + 3 = 6 $
Step 3: Final Answer:
$ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 6 $
Example 90
Problem:
Find $ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 1}{x + 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 1)(x^2 – x + 1)}{x + 1} $
Cancel the common factor:
$ x^2 – x + 1 $
Substitute $ x = -1 $ into the simplified expression:
$ f(-1) = (-1)^2 – (-1) + 1 = 1 + 1 + 1 = 3 $
Step 3: Final Answer:
$ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} = 3 $
Example 91
Problem:
Find $ \lim_{x \to 1} \frac{2x^2 – x}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^2 – x}{x – 1} $
Step 2: Solution:
Substitute $ x = 1 $ into the function:
$ f(1) = \frac{2(1)^2 – (1)}{1 – 1} = \frac{2 – 1}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to 1} \frac{2x^2 – x}{x – 1} $ does not exist.
Example 92
Problem:
Find $ \lim_{x \to 0} \frac{x^3}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3}{x + 1} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{0^3}{0 + 1} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{x^3}{x + 1} = 0 $
Example 93
Problem:
Find $ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 8}{x – 2} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 2)(x^2 + 2x + 4)}{x – 2} $
Cancel the common factor:
$ x^2 + 2x + 4 $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = 2^2 + 2(2) + 4 = 12 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^3 – 8}{x – 2} = 12 $
Example 94
Problem:
Find $ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 1}{x – 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x – 1)(x + 1)}{x – 1} $
Cancel the common factor:
$ x + 1 $
Substitute $ x = 1 $ into the simplified expression:
$ f(1) = 1 + 1 = 2 $
Step 3: Final Answer:
$ \lim_{x \to 1} \frac{x^2 – 1}{x – 1} = 2 $
Example 95
Problem:
Find $ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 1}{x + 1} $
Step 2: Solution:
Factor the numerator:
$ \frac{(x + 1)(x^2 – x + 1)}{x + 1} $
Cancel the common factor:
$ x^2 – x + 1 $
Substitute $ x = -1 $ into the simplified expression:
$ f(-1) = (-1)^2 – (-1) + 1 = 3 $
Step 3: Final Answer:
$ \lim_{x \to -1} \frac{x^3 + 1}{x + 1} = 3 $
Example 96
Problem:
Find $ \lim_{x \to 2} \frac{x^3 – 8}{x^2 – 4} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 – 8}{x^2 – 4} $
Step 2: Solution:
Factor both the numerator and denominator:
$ \frac{(x – 2)(x^2 + 2x + 4)}{(x – 2)(x + 2)} $
Cancel the common factor:
$ \frac{x^2 + 2x + 4}{x + 2} $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = \frac{4 + 4 + 4}{2 + 2} = 3 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^3 – 8}{x^2 – 4} = 3 $
Example 97
Problem:
Find $ \lim_{x \to 0} \frac{2x^3 + 3x^2}{x + 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{2x^3 + 3x^2}{x + 2} $
Step 2: Solution:
Substitute $ x = 0 $ into the function:
$ f(0) = \frac{2(0)^3 + 3(0)^2}{0 + 2} = 0 $
Step 3: Final Answer:
$ \lim_{x \to 0} \frac{2x^3 + 3x^2}{x + 2} = 0 $
Example 98
Problem:
Find $ \lim_{x \to -2} \frac{x^3 + 8}{x – 2} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^3 + 8}{x – 2} $
Step 2: Solution:
Substitute $ x = -2 $ into the function:
$ f(-2) = \frac{(-2)^3 + 8}{-2 – 2} = \frac{-8 + 8}{-4} = \frac{0}{-4} = 0 $
Step 3: Final Answer:
$ \lim_{x \to -2} \frac{x^3 + 8}{x – 2} = 0 $
Example 99
Problem:
Find $ \lim_{x \to 2} \frac{x^2 – 4}{x^2 – 4x} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 – 4}{x^2 – 4x} $
Step 2: Solution:
Factor both the numerator and denominator:
$ \frac{(x – 2)(x + 2)}{x(x – 2)} $
Cancel the common factor:
$ \frac{x + 2}{x} $
Substitute $ x = 2 $ into the simplified expression:
$ f(2) = \frac{2 + 2}{2} = \frac{4}{2} = 2 $
Step 3: Final Answer:
$ \lim_{x \to 2} \frac{x^2 – 4}{x^2 – 4x} = 2 $
Example 100
Problem:
Find $ \lim_{x \to -1} \frac{x^2 + 1}{x + 1} $.
Answer:
Step 1: Given Data:
$ f(x) = \frac{x^2 + 1}{x + 1} $
Step 2: Solution:
Substitute $ x = -1 $ into the function:
$ f(-1) = \frac{(-1)^2 + 1}{-1 + 1} = \frac{1 + 1}{0} $
Step 3: Final Answer:
Since dividing by zero is undefined, $ \lim_{x \to -1} \frac{x^2 + 1}{x + 1} $ does not exist.