Evaluating Limits Worksheet Solutions - PreCalculus | Exercises ... - Free Printable

Problem: Evaluate the following limits without using a calculator.

We will solve each limit step by step, explaining the methods used.

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#### 1. $\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3}$

Solution:

1. Factor the numerator:
The numerator is $2x^2 - 5x - 3$. We factor it as follows:
\[
2x^2 - 5x - 3 = (2x + 1)(x - 3)
\]
This can be verified by expanding $(2x + 1)(x - 3)$:
\[
(2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3
\]

2. Simplify the expression:
Substitute the factored form into the limit:
\[
\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3} = \lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3}
\]
Cancel the common factor $(x - 3)$ (since $x \neq 3$):
\[
\lim_{x \to 3} (2x + 1)
\]

3. Evaluate the limit:
Substitute $x = 3$ into $2x + 1$:
\[
2(3) + 1 = 6 + 1 = 7
\]

Final Answer:
\[
\boxed{7}
\]

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#### 2. $\lim_{x \to 2} \frac{x^4 - 16}{x - 2}$

Solution:

1. Factor the numerator:
The numerator is $x^4 - 16$, which is a difference of squares:
\[
x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4)
\]
Further factor $x^2 - 4$ (another difference of squares):
\[
x^2 - 4 = (x - 2)(x + 2)
\]
So, the full factorization is:
\[
x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)
\]

2. Simplify the expression:
Substitute the factored form into the limit:
\[
\lim_{x \to 2} \frac{x^4 - 16}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)(x^2 + 4)}{x - 2}
\]
Cancel the common factor $(x - 2)$ (since $x \neq 2$):
\[
\lim_{x \to 2} (x + 2)(x^2 + 4)
\]

3. Evaluate the limit:
Substitute $x = 2$ into $(x + 2)(x^2 + 4)$:
\[
(2 + 2)(2^2 + 4) = 4(4 + 4) = 4 \cdot 8 = 32
\]

Final Answer:
\[
\boxed{32}
\]

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#### 3. $\lim_{x \to -1} \frac{x^4 + 3x^3 - x^2 + x + 4}{x + 1}$

Solution:

1. Use polynomial long division:
Divide $x^4 + 3x^3 - x^2 + x + 4$ by $x + 1$.

- Step 1: Divide the leading term $x^4$ by $x$ to get $x^3$.
- Step 2: Multiply $x^3$ by $x + 1$ to get $x^4 + x^3$.
- Step 3: Subtract $x^4 + x^3$ from $x^4 + 3x^3 - x^2 + x + 4$ to get $2x^3 - x^2 + x + 4$.
- Step 4: Repeat the process with $2x^3$: divide $2x^3$ by $x$ to get $2x^2$.
- Step 5: Multiply $2x^2$ by $x + 1$ to get $2x^3 + 2x^2$.
- Step 6: Subtract $2x^3 + 2x^2$ from $2x^3 - x^2 + x + 4$ to get $-3x^2 + x + 4$.
- Step 7: Repeat the process with $-3x^2$: divide $-3x^2$ by $x$ to get $-3x$.
- Step 8: Multiply $-3x$ by $x + 1$ to get $-3x^2 - 3x$.
- Step 9: Subtract $-3x^2 - 3x$ from $-3x^2 + x + 4$ to get $4x + 4$.
- Step 10: Repeat the process with $4x$: divide $4x$ by $x$ to get $4$.
- Step 11: Multiply $4$ by $x + 1$ to get $4x + 4$.
- Step 12: Subtract $4x + 4$ from $4x + 4$ to get $0$.

The quotient is $x^3 + 2x^2 - 3x + 4$.

2. Simplify the expression:
\[
\lim_{x \to -1} \frac{x^4 + 3x^3 - x^2 + x + 4}{x + 1} = \lim_{x \to -1} (x^3 + 2x^2 - 3x + 4)
\]

3. Evaluate the limit:
Substitute $x = -1$ into $x^3 + 2x^2 - 3x + 4$:
\[
(-1)^3 + 2(-1)^2 - 3(-1) + 4 = -1 + 2 + 3 + 4 = 8
\]

Final Answer:
\[
\boxed{8}
\]

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#### 4. $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$

Solution:

1. Rationalize the numerator:
Multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{x + 4} + 2$:
\[
\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}
\]
Simplify the numerator using the difference of squares:
\[
(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2) = (\sqrt{x + 4})^2 - 2^2 = (x + 4) - 4 = x
\]
So the expression becomes:
\[
\lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)}
\]

2. Simplify the expression:
Cancel the common factor $x$ (since $x \neq 0$):
\[
\lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}
\]

3. Evaluate the limit:
Substitute $x = 0$ into $\frac{1}{\sqrt{x + 4} + 2}$:
\[
\frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
\]

Final Answer:
\[
\boxed{\frac{1}{4}}
\]

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#### 5. $\lim_{x \to 3} \frac{\sqrt{x + 6} - x}{x - 3}$

Solution:

1. Rationalize the numerator:
Multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{x + 6} + x$:
\[
\lim_{x \to 3} \frac{\sqrt{x + 6} - x}{x - 3} \cdot \frac{\sqrt{x + 6} + x}{\sqrt{x + 6} + x}
\]
Simplify the numerator using the difference of squares:
\[
(\sqrt{x + 6} - x)(\sqrt{x + 6} + x) = (\sqrt{x + 6})^2 - x^2 = (x + 6) - x^2 = -x^2 + x + 6
\]
So the expression becomes:
\[
\lim_{x \to 3} \frac{-x^2 + x + 6}{(x - 3)(\sqrt{x + 6} + x)}
\]

2. Factor the numerator:
Factor $-x^2 + x + 6$:
\[
-x^2 + x + 6 = -(x^2 - x - 6) = -(x - 3)(x + 2)
\]
So the expression becomes:
\[
\lim_{x \to 3} \frac{-(x - 3)(x + 2)}{(x - 3)(\sqrt{x + 6} + x)}
\]

3. Simplify the expression:
Cancel the common factor $(x - 3)$ (since $x \neq 3$):
\[
\lim_{x \to 3} \frac{-(x + 2)}{\sqrt{x + 6} + x}
\]

4. Evaluate the limit:
Substitute $x = 3$ into $\frac{-(x + 2)}{\sqrt{x + 6} + x}$:
\[
\frac{-(3 + 2)}{\sqrt{3 + 6} + 3} = \frac{-5}{\sqrt{9} + 3} = \frac{-5}{3 + 3} = \frac{-5}{6}
\]

Final Answer:
\[
\boxed{-\frac{5}{6}}
\]

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Final Answers:

1. $\boxed{7}$
2. $\boxed{32}$
3. $\boxed{8}$
4. $\boxed{\frac{1}{4}}$
5. $\boxed{-\frac{5}{6}}$