Comprehensive practice sheet covering various types of limit problems found in introductory calculus courses.

Worksheet: Limits

#### Section A: Evaluate

1. Evaluate \( \lim_{x \to 2} (x^3 - 2x + 1) \)

This is a polynomial function, and polynomials are continuous everywhere. Therefore, we can directly substitute \( x = 2 \):

\[
\lim_{x \to 2} (x^3 - 2x + 1) = 2^3 - 2(2) + 1 = 8 - 4 + 1 = 5
\]

Answer: \( \boxed{5} \)

2. Evaluate \( \lim_{x \to 0} \frac{x^2 - 2x + 3}{x^2 + 4x + 3} \)

Substitute \( x = 0 \) into the expression:

\[
\lim_{x \to 0} \frac{x^2 - 2x + 3}{x^2 + 4x + 3} = \frac{0^2 - 2(0) + 3}{0^2 + 4(0) + 3} = \frac{3}{3} = 1
\]

Answer: \( \boxed{1} \)

3. Evaluate \( \lim_{x \to 1} \frac{x^3 + x}{x^3 + x - 2} \)

Substitute \( x = 1 \) into the expression:

\[
\lim_{x \to 1} \frac{x^3 + x}{x^3 + x - 2} = \frac{1^3 + 1}{1^3 + 1 - 2} = \frac{1 + 1}{1 + 1 - 2} = \frac{2}{0}
\]

The denominator is zero, and the numerator is non-zero, so the limit does not exist (it approaches infinity).

Answer: \( \boxed{\text{DNE}} \)

4. Evaluate \( \lim_{x \to 1} \frac{x - 1}{x^2 + x - 2} \)

Factor the denominator:

\[
x^2 + x - 2 = (x - 1)(x + 2)
\]

So the expression becomes:

\[
\lim_{x \to 1} \frac{x - 1}{(x - 1)(x + 2)}
\]

Cancel \( x - 1 \) (since \( x \neq 1 \)):

\[
\lim_{x \to 1} \frac{1}{x + 2} = \frac{1}{1 + 2} = \frac{1}{3}
\]

Answer: \( \boxed{\frac{1}{3}} \)

5. Evaluate \( \lim_{x \to 2} \frac{x^3 - 4}{x^2 + x - 2} \)

Factor the denominator:

\[
x^2 + x - 2 = (x - 1)(x + 2)
\]

The numerator \( x^3 - 4 \) cannot be factored easily, so we substitute \( x = 2 \):

\[
\lim_{x \to 2} \frac{x^3 - 4}{x^2 + x - 2} = \frac{2^3 - 4}{2^2 + 2 - 2} = \frac{8 - 4}{4 + 2 - 2} = \frac{4}{4} = 1
\]

Answer: \( \boxed{1} \)

6. Evaluate \( \lim_{x \to -1} \frac{x^3 - 1}{x^2 + x - 2} \)

Factor the denominator:

\[
x^2 + x - 2 = (x - 1)(x + 2)
\]

The numerator \( x^3 - 1 \) can be factored using the difference of cubes:

\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]

So the expression becomes:

\[
\lim_{x \to -1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 2)}
\]

Cancel \( x - 1 \) (since \( x \neq 1 \)):

\[
\lim_{x \to -1} \frac{x^2 + x + 1}{x + 2} = \frac{(-1)^2 + (-1) + 1}{-1 + 2} = \frac{1 - 1 + 1}{1} = 1
\]

Answer: \( \boxed{1} \)

7. Evaluate \( \lim_{x \to 1} \frac{x^2 - 1}{x^2 - 2x + 1} \)

Factor both the numerator and the denominator:

\[
x^2 - 1 = (x - 1)(x + 1)
\]
\[
x^2 - 2x + 1 = (x - 1)^2
\]

So the expression becomes:

\[
\lim_{x \to 1} \frac{(x - 1)(x + 1)}{(x - 1)^2} = \lim_{x \to 1} \frac{x + 1}{x - 1}
\]

As \( x \to 1 \), the denominator approaches zero, and the numerator approaches 2. The limit does not exist (it approaches infinity).

Answer: \( \boxed{\text{DNE}} \)

8. Evaluate \( \lim_{x \to 1} \frac{x^2 - 2x + 1}{x^2 + 3x + 2} \)

Factor both the numerator and the denominator:

\[
x^2 - 2x + 1 = (x - 1)^2
\]
\[
x^2 + 3x + 2 = (x + 1)(x + 2)
\]

So the expression becomes:

\[
\lim_{x \to 1} \frac{(x - 1)^2}{(x + 1)(x + 2)}
\]

Substitute \( x = 1 \):

\[
\lim_{x \to 1} \frac{(1 - 1)^2}{(1 + 1)(1 + 2)} = \frac{0}{2 \cdot 3} = 0
\]

Answer: \( \boxed{0} \)

9. Evaluate \( \lim_{x \to 0} \frac{x^2 - x + 6}{x^3 - x} \)

Factor the denominator:

\[
x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)
\]

The numerator \( x^2 - x + 6 \) cannot be factored easily, so we substitute \( x = 0 \):

\[
\lim_{x \to 0} \frac{x^2 - x + 6}{x^3 - x} = \frac{0^2 - 0 + 6}{0^3 - 0} = \frac{6}{0}
\]

The denominator is zero, and the numerator is non-zero, so the limit does not exist (it approaches infinity).

Answer: \( \boxed{\text{DNE}} \)

---

Section B: Evaluate

1. Evaluate \( \lim_{x \to 1} (x^2 - x + \sqrt{2x - 1}) \)

Substitute \( x = 1 \):

\[
\lim_{x \to 1} (x^2 - x + \sqrt{2x - 1}) = 1^2 - 1 + \sqrt{2(1) - 1} = 1 - 1 + \sqrt{1} = 0 + 1 = 1
\]

Answer: \( \boxed{1} \)

2. Evaluate \( \lim_{x \to 3} \sqrt{2x + 3} \)

Substitute \( x = 3 \):

\[
\lim_{x \to 3} \sqrt{2x + 3} = \sqrt{2(3) + 3} = \sqrt{6 + 3} = \sqrt{9} = 3
\]

Answer: \( \boxed{3} \)

3. Evaluate \( \lim_{x \to 0} \frac{\sqrt{1 + x} + \sqrt{1 - x}}{x} \)

This is an indeterminate form \( \frac{0}{0} \). Use L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{\sqrt{1 + x} + \sqrt{1 - x}}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sqrt{1 + x} + \sqrt{1 - x})}{\frac{d}{dx}(x)}
\]

Differentiate the numerator and the denominator:

\[
\frac{d}{dx}(\sqrt{1 + x}) = \frac{1}{2\sqrt{1 + x}}, \quad \frac{d}{dx}(\sqrt{1 - x}) = \frac{-1}{2\sqrt{1 - x}}
\]

So:

\[
\lim_{x \to 0} \frac{\frac{1}{2\sqrt{1 + x}} - \frac{1}{2\sqrt{1 - x}}}{1} = \frac{1}{2\sqrt{1 + 0}} - \frac{1}{2\sqrt{1 - 0}} = \frac{1}{2} - \frac{1}{2} = 0
\]

Answer: \( \boxed{0} \)

4. Evaluate \( \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \)

Rationalize the numerator by multiplying by the conjugate:

\[
\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \cdot \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} = \lim_{x \to 0} \frac{(1 + x) - (1 - x)}{(\sqrt{1 + x} + \sqrt{1 - x})^2}
\]

Simplify the numerator:

\[
(1 + x) - (1 - x) = 2x
\]

So:

\[
\lim_{x \to 0} \frac{2x}{(\sqrt{1 + x} + \sqrt{1 - x})^2}
\]

Substitute \( x = 0 \):

\[
\frac{2(0)}{(\sqrt{1 + 0} + \sqrt{1 - 0})^2} = \frac{0}{(1 + 1)^2} = \frac{0}{4} = 0
\]

Answer: \( \boxed{0} \)

5. Evaluate \( \lim_{x \to 0} \frac{\sqrt{9 + x} - 3}{x} \)

This is an indeterminate form \( \frac{0}{0} \). Use L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{\sqrt{9 + x} - 3}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sqrt{9 + x} - 3)}{\frac{d}{dx}(x)}
\]

Differentiate the numerator and the denominator:

\[
\frac{d}{dx}(\sqrt{9 + x}) = \frac{1}{2\sqrt{9 + x}}, \quad \frac{d}{dx}(3) = 0
\]

So:

\[
\lim_{x \to 0} \frac{\frac{1}{2\sqrt{9 + x}}}{1} = \frac{1}{2\sqrt{9 + 0}} = \frac{1}{2 \cdot 3} = \frac{1}{6}
\]

Answer: \( \boxed{\frac{1}{6}} \)

6. Evaluate \( \lim_{x \to 1} \frac{\sqrt{1 - x}}{x - 1} \)

Let \( u = 1 - x \). As \( x \to 1 \), \( u \to 0 \). Rewrite the limit in terms of \( u \):

\[
\lim_{x \to 1} \frac{\sqrt{1 - x}}{x - 1} = \lim_{u \to 0} \frac{\sqrt{u}}{-u} = \lim_{u \to 0} \frac{1}{-\sqrt{u}} = -\infty
\]

Answer: \( \boxed{-\infty} \)

7. Evaluate \( \lim_{x \to 0} \frac{\sqrt{1 + 2x} - \sqrt{1 - 3x}}{x} \)

This is an indeterminate form \( \frac{0}{0} \). Use L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{\sqrt{1 + 2x} - \sqrt{1 - 3x}}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sqrt{1 + 2x} - \sqrt{1 - 3x})}{\frac{d}{dx}(x)}
\]

Differentiate the numerator and the denominator:

\[
\frac{d}{dx}(\sqrt{1 + 2x}) = \frac{2}{2\sqrt{1 + 2x}}, \quad \frac{d}{dx}(\sqrt{1 - 3x}) = \frac{-3}{2\sqrt{1 - 3x}}
\]

So:

\[
\lim_{x \to 0} \frac{\frac{2}{2\sqrt{1 + 2x}} - \frac{-3}{2\sqrt{1 - 3x}}}{1} = \lim_{x \to 0} \left( \frac{1}{\sqrt{1 + 2x}} + \frac{3}{2\sqrt{1 - 3x}} \right)
\]

Substitute \( x = 0 \):

\[
\frac{1}{\sqrt{1 + 2(0)}} + \frac{3}{2\sqrt{1 - 3(0)}} = \frac{1}{\sqrt{1}} + \frac{3}{2\sqrt{1}} = 1 + \frac{3}{2} = \frac{5}{2}
\]

Answer: \( \boxed{\frac{5}{2}} \)

8. Evaluate \( \lim_{x \to 0} \frac{x}{\sqrt{x^2 + 2x} - x} \)

Rationalize the denominator by multiplying by the conjugate:

\[
\lim_{x \to 0} \frac{x}{\sqrt{x^2 + 2x} - x} \cdot \frac{\sqrt{x^2 + 2x} + x}{\sqrt{x^2 + 2x} + x} = \lim_{x \to 0} \frac{x(\sqrt{x^2 + 2x} + x)}{(x^2 + 2x) - x^2}
\]

Simplify the denominator:

\[
(x^2 + 2x) - x^2 = 2x
\]

So:

\[
\lim_{x \to 0} \frac{x(\sqrt{x^2 + 2x} + x)}{2x} = \lim_{x \to 0} \frac{\sqrt{x^2 + 2x} + x}{2}
\]

Substitute \( x = 0 \):

\[
\frac{\sqrt{0^2 + 2(0)} + 0}{2} = \frac{0 + 0}{2} = 0
\]

Answer: \( \boxed{0} \)

9. Evaluate \( \lim_{x \to 0} \frac{\sqrt{x^3 + 9} - 3}{x} \)

This is an indeterminate form \( \frac{0}{0} \). Use L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{\sqrt{x^3 + 9} - 3}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sqrt{x^3 + 9} - 3)}{\frac{d}{dx}(x)}
\]

Differentiate the numerator and the denominator:

\[
\frac{d}{dx}(\sqrt{x^3 + 9}) = \frac{3x^2}{2\sqrt{x^3 + 9}}, \quad \frac{d}{dx}(3) = 0
\]

So:

\[
\lim_{x \to 0} \frac{\frac{3x^2}{2\sqrt{x^3 + 9}}}{1} = \lim_{x \to 0} \frac{3x^2}{2\sqrt{x^3 + 9}}
\]

Substitute \( x = 0 \):

\[
\frac{3(0)^2}{2\sqrt{0^3 + 9}} = \frac{0}{2\sqrt{9}} = 0
\]

Answer: \( \boxed{0} \)

---

Section C: Evaluate

1. Evaluate \( \lim_{x \to 0} \left( \frac{1}{x} + \frac{1}{x} \right) \)

Simplify the expression:

\[
\frac{1}{x} + \frac{1}{x} = \frac{2}{x}
\]

As \( x \to 0 \), \( \frac{2}{x} \) approaches infinity.

Answer: \( \boxed{\infty} \)

2. Evaluate \( \lim_{x \to 0} \left( \frac{1}{x} + \frac{1}{x^2} \right) \)

Simplify the expression:

\[
\frac{1}{x} + \frac{1}{x^2} = \frac{x + 1}{x^2}
\]

As \( x \to 0 \), \( \frac{x + 1}{x^2} \) approaches infinity.

Answer: \( \boxed{\infty} \)

---

Section D: Evaluate the following one-sided limits

1. Evaluate \( \lim_{x \to 2^-} \frac{1}{x - 2} \)

As \( x \to 2^- \), \( x - 2 \) approaches 0 from the left (negative values). Therefore, \( \frac{1}{x - 2} \) approaches negative infinity.

Answer: \( \boxed{-\infty} \)

2. Evaluate \( \lim_{x \to -2^+} \frac{1}{x + 2} \)

As \( x \to -2^+ \), \( x + 2 \) approaches 0 from the right (positive values). Therefore, \( \frac{1}{x + 2} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

3. Evaluate \( \lim_{x \to 3^-} \frac{x - 1}{x - 3} \)

As \( x \to 3^- \), \( x - 3 \) approaches 0 from the left (negative values). The numerator \( x - 1 \) approaches 2. Therefore, \( \frac{x - 1}{x - 3} \) approaches negative infinity.

Answer: \( \boxed{-\infty} \)

4. Evaluate \( \lim_{x \to 4^-} \frac{x - 1}{4 - x} \)

As \( x \to 4^- \), \( 4 - x \) approaches 0 from the right (positive values). The numerator \( x - 1 \) approaches 3. Therefore, \( \frac{x - 1}{4 - x} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

5. Evaluate \( \lim_{x \to 1^-} \frac{x + 2}{(x + 1)^2} \)

As \( x \to 1^- \), \( x + 1 \) approaches 2 from the left. The numerator \( x + 2 \) approaches 3. Therefore, \( \frac{x + 2}{(x + 1)^2} \) approaches:

\[
\frac{3}{2^2} = \frac{3}{4}
\]

Answer: \( \boxed{\frac{3}{4}} \)

6. Evaluate \( \lim_{x \to 0^+} \frac{(x + 2)(x - 1)}{x^2} \)

As \( x \to 0^+ \), \( x^2 \) approaches 0 from the right (positive values). The numerator \( (x + 2)(x - 1) \) approaches \( (0 + 2)(0 - 1) = -2 \). Therefore, \( \frac{(x + 2)(x - 1)}{x^2} \) approaches negative infinity.

Answer: \( \boxed{-\infty} \)

7. Evaluate \( \lim_{x \to 5^+} \frac{x - 1}{x(x - 5)} \)

As \( x \to 5^+ \), \( x - 5 \) approaches 0 from the right (positive values). The numerator \( x - 1 \) approaches 4. The denominator \( x(x - 5) \) approaches \( 5 \cdot 0^+ = 0^+ \). Therefore, \( \frac{x - 1}{x(x - 5)} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

8. Evaluate \( \lim_{x \to 1^-} \frac{x^2 - x + 1}{x^2 - 2x + 1} \)

Factor the denominator:

\[
x^2 - 2x + 1 = (x - 1)^2
\]

So the expression becomes:

\[
\lim_{x \to 1^-} \frac{x^2 - x + 1}{(x - 1)^2}
\]

As \( x \to 1^- \), \( (x - 1)^2 \) approaches 0 from the right (positive values). The numerator \( x^2 - x + 1 \) approaches \( 1^2 - 1 + 1 = 1 \). Therefore, \( \frac{x^2 - x + 1}{(x - 1)^2} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

9. Evaluate \( \lim_{x \to 1^+} \frac{x^2 + 8x - 9}{x^2 - 2x + 1} \)

Factor both the numerator and the denominator:

\[
x^2 + 8x - 9 = (x - 1)(x + 9)
\]
\[
x^2 - 2x + 1 = (x - 1)^2
\]

So the expression becomes:

\[
\lim_{x \to 1^+} \frac{(x - 1)(x + 9)}{(x - 1)^2} = \lim_{x \to 1^+} \frac{x + 9}{x - 1}
\]

As \( x \to 1^+ \), \( x - 1 \) approaches 0 from the right (positive values). The numerator \( x + 9 \) approaches 10. Therefore, \( \frac{x + 9}{x - 1} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

---

Section E: Evaluate

1. Evaluate \( \lim_{x \to 1^-} \frac{1}{\sqrt{1 - x}} \)

As \( x \to 1^- \), \( 1 - x \) approaches 0 from the right (positive values). Therefore, \( \sqrt{1 - x} \) approaches 0 from the right, and \( \frac{1}{\sqrt{1 - x}} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

2. Evaluate \( \lim_{x \to 2^-} \frac{1 - x}{\sqrt{x^2 - 4}} \)

As \( x \to 2^- \), \( x^2 - 4 \) approaches 0 from the left (negative values). However, the square root function is not defined for negative values. Therefore, the limit does not exist.

Answer: \( \boxed{\text{DNE}} \)

3. Evaluate \( \lim_{x \to 4^-} \frac{\sqrt{16 - x^2}}{x - 4} \)

As \( x \to 4^- \), \( x - 4 \) approaches 0 from the left (negative values). The numerator \( \sqrt{16 - x^2} \) approaches 0 from the right (positive values). Therefore, \( \frac{\sqrt{16 - x^2}}{x - 4} \) approaches negative infinity.

Answer: \( \boxed{-\infty} \)

4. Evaluate \( \lim_{x \to 3^+} \frac{\sqrt{x^2 - 9}}{x - 3} \)

As \( x \to 3^+ \), \( x - 3 \) approaches 0 from the right (positive values). The numerator \( \sqrt{x^2 - 9} \) approaches 0 from the right (positive values). Therefore, \( \frac{\sqrt{x^2 - 9}}{x - 3} \) approaches positive infinity.

Answer: \( \boxed{\infty} \)

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{A1: } 5 & \text{A2: } 1 & \text{A3: } \text{DNE} \\
\text{A4: } \frac{1}{3} & \text{A5: } 1 & \text{A6: } 1 \\
\text{A7: } \text{DNE} & \text{A8: } 0 & \text{A9: } \text{DNE} \\
\text{B1: } 1 & \text{B2: } 3 & \text{B3: } 0 \\
\text{B4: } 0 & \text{B5: } \frac{1}{6} & \text{B6: } -\infty \\
\text{B7: } \frac{5}{2} & \text{B8: } 0 & \text{B9: } 0 \\
\text{C1: } \infty & \text{C2: } \infty \\
\text{D1: } -\infty & \text{D2: } \infty & \text{D3: } -\infty \\
\text{D4: } \infty & \text{D5: } \frac{3}{4} & \text{D6: } -\infty \\
\text{D7: } \infty & \text{D8: } \infty & \text{D9: } \infty \\
\text{E1: } \infty & \text{E2: } \text{DNE} & \text{E3: } -\infty \\
\text{E4: } \infty \\
\end{array}
}
\]