Solved WORKSHEET - Limits A) Evaluate: 1) limx2x+1 r2x 3 2 ... - Free Printable

Problem: Evaluate the given limits.

The worksheet is divided into several sections (A, B, C, D, E). Below, I will solve each part step by step.

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Section A: Evaluate

#### 1. $\lim_{x \to 2} \frac{x^3 - 2x + 1}{x^2 + x - 6}$

- Step 1: Factor the denominator.
$$
x^2 + x - 6 = (x - 2)(x + 3)
$$

- Step 2: Check if the numerator can be factored or simplified at $x = 2$.
$$
x^3 - 2x + 1
$$
Direct substitution of $x = 2$:
$$
2^3 - 2(2) + 1 = 8 - 4 + 1 = 5
$$
Since the numerator does not equal zero when $x = 2$, we can directly substitute $x = 2$ into the limit.

- Step 3: Substitute $x = 2$ into the expression.
$$
\lim_{x \to 2} \frac{x^3 - 2x + 1}{x^2 + x - 6} = \frac{5}{(2 - 2)(2 + 3)} = \frac{5}{0}
$$
The denominator becomes zero, and the numerator is non-zero, so the limit does not exist (it approaches infinity).

Answer: $\boxed{\text{DNE}}$

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

- Step 1: Factor both the numerator and the denominator.
$$
x^2 - 2x - 3 = (x - 3)(x + 1)
$$
$$
x^2 + 4x + 3 = (x + 1)(x + 3)
$$

- Step 2: Simplify the fraction.
$$
\frac{x^2 - 2x - 3}{x^2 + 4x + 3} = \frac{(x - 3)(x + 1)}{(x + 1)(x + 3)}
$$
Cancel the common factor $(x + 1)$ (valid for $x \neq -1$):
$$
\frac{x - 3}{x + 3}
$$

- Step 3: Substitute $x = -1$ into the simplified expression.
$$
\lim_{x \to -1} \frac{x - 3}{x + 3} = \frac{-1 - 3}{-1 + 3} = \frac{-4}{2} = -2
$$

Answer: $\boxed{-2}$

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

- Step 1: Factor the denominator.
$$
x^2 - x - 2 = (x - 2)(x + 1)
$$

- Step 2: Check if the numerator can be factored or simplified at $x = 2$.
$$
x^2 + x = x(x + 1)
$$
Direct substitution of $x = 2$:
$$
2^2 + 2 = 4 + 2 = 6
$$
Since the numerator does not equal zero when $x = 2$, we can directly substitute $x = 2$ into the limit.

- Step 3: Substitute $x = 2$ into the expression.
$$
\lim_{x \to 2} \frac{x^2 + x}{x^2 - x - 2} = \frac{6}{(2 - 2)(2 + 1)} = \frac{6}{0}
$$
The denominator becomes zero, and the numerator is non-zero, so the limit does not exist (it approaches infinity).

Answer: $\boxed{\text{DNE}}$

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

- Step 1: Divide the numerator and the denominator by the highest power of $x$ in the denominator ($x^2$).
$$
\frac{x - 1}{x^2 + x - 2} = \frac{\frac{x}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} + \frac{x}{x^2} - \frac{2}{x^2}} = \frac{\frac{1}{x} - \frac{1}{x^2}}{1 + \frac{1}{x} - \frac{2}{x^2}}
$$

- Step 2: Take the limit as $x \to \infty$.
$$
\lim_{x \to \infty} \frac{\frac{1}{x} - \frac{1}{x^2}}{1 + \frac{1}{x} - \frac{2}{x^2}} = \frac{0 - 0}{1 + 0 - 0} = \frac{0}{1} = 0
$$

Answer: $\boxed{0}$

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

- Step 1: Factor both the numerator and the denominator.
$$
x^2 - 4 = (x - 2)(x + 2)
$$
$$
x^2 - x - 2 = (x - 2)(x + 1)
$$

- Step 2: Simplify the fraction.
$$
\frac{x^2 - 4}{x^2 - x - 2} = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)}
$$
Cancel the common factor $(x - 2)$ (valid for $x \neq 2$):
$$
\frac{x + 2}{x + 1}
$$

- Step 3: Substitute $x = 2$ into the simplified expression.
$$
\lim_{x \to 2} \frac{x + 2}{x + 1} = \frac{2 + 2}{2 + 1} = \frac{4}{3}
$$

Answer: $\boxed{\frac{4}{3}}$

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

- Step 1: Factor both the numerator and the denominator.
$$
x^2 - 1 = (x - 1)(x + 1)
$$
$$
x^2 + x - 2 = (x - 1)(x + 2)
$$

- Step 2: Simplify the fraction.
$$
\frac{x^2 - 1}{x^2 + x - 2} = \frac{(x - 1)(x + 1)}{(x - 1)(x + 2)}
$$
Cancel the common factor $(x - 1)$ (valid for $x \neq 1$):
$$
\frac{x + 1}{x + 2}
$$

- Step 3: Substitute $x = -1$ into the simplified expression.
$$
\lim_{x \to -1} \frac{x + 1}{x + 2} = \frac{-1 + 1}{-1 + 2} = \frac{0}{1} = 0
$$

Answer: $\boxed{0}$

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

- Step 1: Divide the numerator and the denominator by the highest power of $x$ in the denominator ($x^3$).
$$
\frac{x^2 - 1}{x^3 - 2x + 1} = \frac{\frac{x^2}{x^3} - \frac{1}{x^3}}{\frac{x^3}{x^3} - \frac{2x}{x^3} + \frac{1}{x^3}} = \frac{\frac{1}{x} - \frac{1}{x^3}}{1 - \frac{2}{x^2} + \frac{1}{x^3}}
$$

- Step 2: Take the limit as $x \to \infty$.
$$
\lim_{x \to \infty} \frac{\frac{1}{x} - \frac{1}{x^3}}{1 - \frac{2}{x^2} + \frac{1}{x^3}} = \frac{0 - 0}{1 - 0 + 0} = \frac{0}{1} = 0
$$

Answer: $\boxed{0}$

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

- Step 1: Divide the numerator and the denominator by the highest power of $x$ in the denominator ($x^3$).
$$
\frac{x^2 - 2x + 1}{x^3 + 3x + 2} = \frac{\frac{x^2}{x^3} - \frac{2x}{x^3} + \frac{1}{x^3}}{\frac{x^3}{x^3} + \frac{3x}{x^3} + \frac{2}{x^3}} = \frac{\frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^3}}{1 + \frac{3}{x^2} + \frac{2}{x^3}}
$$

- Step 2: Take the limit as $x \to \infty$.
$$
\lim_{x \to \infty} \frac{\frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^3}}{1 + \frac{3}{x^2} + \frac{2}{x^3}} = \frac{0 - 0 + 0}{1 + 0 + 0} = \frac{0}{1} = 0
$$

Answer: $\boxed{0}$

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#### 9. $\lim_{x \to 2} \frac{x^3 + x + 6}{x^3 - x}$

- Step 1: Factor the denominator.
$$
x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)
$$

- Step 2: Check if the numerator can be factored or simplified at $x = 2$.
$$
x^3 + x + 6
$$
Direct substitution of $x = 2$:
$$
2^3 + 2 + 6 = 8 + 2 + 6 = 16
$$
Since the numerator does not equal zero when $x = 2$, we can directly substitute $x = 2$ into the limit.

- Step 3: Substitute $x = 2$ into the expression.
$$
\lim_{x \to 2} \frac{x^3 + x + 6}{x^3 - x} = \frac{16}{2(2 - 1)(2 + 1)} = \frac{16}{2 \cdot 1 \cdot 3} = \frac{16}{6} = \frac{8}{3}
$$

Answer: $\boxed{\frac{8}{3}}$

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Final Answers for Section A:

1. $\boxed{\text{DNE}}$
2. $\boxed{-2}$
3. $\boxed{\text{DNE}}$
4. $\boxed{0}$
5. $\boxed{\frac{4}{3}}$
6. $\boxed{0}$
7. $\boxed{0}$
8. $\boxed{0}$
9. $\boxed{\frac{8}{3}}$

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Continue solving other sections similarly. Let me know if you need further assistance!