Math worksheet featuring limit problems from exercises 15 to 32, including algebraic and trigonometric functions.

Let's solve each of the limit problems from Exercises 15–32 step by step. I'll go through them one by one, explaining the reasoning and applying appropriate limit techniques.

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Exercises 15–18: Find each of the limits, if possible

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#### 15.
(a) $\lim_{x \to 0} \frac{x^2 + 2}{x^3 - 1}$

- Plug in $x = 0$:
Numerator: $0^2 + 2 = 2$
Denominator: $0^3 - 1 = -1$
So, $\frac{2}{-1} = -2$

✔ Answer: $-2$

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(b) $\lim_{x \to \infty} \frac{x^2 + 2}{x^2 - 1}$

- Divide numerator and denominator by $x^2$:
$$
\frac{1 + \frac{2}{x^2}}{1 - \frac{1}{x^2}}
$$
- As $x \to \infty$, $\frac{2}{x^2} \to 0$, $\frac{1}{x^2} \to 0$
So, limit is $\frac{1 + 0}{1 - 0} = 1$

✔ Answer: $1$

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(c) $\lim_{x \to \infty} \frac{x^2 + 2}{x - 1}$

- Degree of numerator (2) > degree of denominator (1), so limit is $\infty$
But let’s analyze:
$$
\frac{x^2 + 2}{x - 1} \approx \frac{x^2}{x} = x \to \infty
$$
Since numerator grows faster than denominator, and both positive for large $x$, the limit is $+\infty$

✔ Answer: $\infty$

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#### 16.
(a) $\lim_{x \to \infty} \frac{3 - 2x}{3x^2 - 1}$

- Degree of numerator: 1, denominator: 2 → denominator dominates
- Divide numerator and denominator by $x^2$:
$$
\frac{\frac{3}{x^2} - \frac{2}{x}}{3 - \frac{1}{x^2}} \to \frac{0 - 0}{3 - 0} = 0
$$

✔ Answer: $0$

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(b) $\lim_{x \to \infty} \frac{3 - 2x}{3x - 1}$

- Degrees are equal (both 1), so divide by $x$:
$$
\frac{\frac{3}{x} - 2}{3 - \frac{1}{x}} \to \frac{0 - 2}{3 - 0} = -\frac{2}{3}
$$

✔ Answer: $-\frac{2}{3}$

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(c) $\lim_{x \to \infty} \frac{3 - 2x^2}{3x - 1}$

- Numerator degree: 2, denominator: 1 → numerator dominates
- As $x \to \infty$, $-2x^2$ dominates → negative infinity
So, limit is $-\infty$

✔ Answer: $-\infty$

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#### 17.
(a) $\lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x^2 - 4}$

- Highest power: $x^{3/2}$ in numerator, $x^2$ in denominator
- Since $x^2$ grows faster than $x^{3/2}$, denominator dominates
- Divide numerator and denominator by $x^2$:
$$
\frac{\frac{5}{x^2} - 2x^{-1/2}}{1 - \frac{4}{x^2}} \to \frac{0 - 0}{1 - 0} = 0
$$

✔ Answer: $0$

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(b) $\lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x^{3/2} - 4}$

- Both numerator and denominator have $x^{3/2}$ as dominant term
- Divide numerator and denominator by $x^{3/2}$:
$$
\frac{\frac{5}{x^{3/2}} - 2}{3 - \frac{4}{x^{3/2}}} \to \frac{0 - 2}{3 - 0} = -\frac{2}{3}
$$

✔ Answer: $-\frac{2}{3}$

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(c) $\lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x - 4}$

- Numerator has $x^{3/2}$, denominator has $x$
- $x^{3/2} = x \cdot \sqrt{x}$ grows faster than $x$
- So numerator dominates → $-\infty$ (since $-2x^{3/2}$ dominates)

✔ Answer: $-\infty$

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#### 18.
(a) $\lim_{x \to \infty} \frac{5x^{3/2}}{4x^2 + 1}$

- Numerator: $x^{3/2}$, denominator: $x^2$
- $x^2$ grows faster → denominator dominates
- Divide by $x^2$:
$$
\frac{5x^{-1/2}}{1 + \frac{1}{x^2}} \to \frac{0}{1} = 0
$$

✔ Answer: $0$

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(b) $\lim_{x \to \infty} \frac{5x^{3/2}}{4x^{3/2} + 1}$

- Same degree: $x^{3/2}$
- Divide by $x^{3/2}$:
$$
\frac{5}{4 + \frac{1}{x^{3/2}}} \to \frac{5}{4 + 0} = \frac{5}{4}
$$

✔ Answer: $\frac{5}{4}$

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(c) $\lim_{x \to \infty} \frac{5x^{3/2}}{4\sqrt{x} + 1}$

- Note: $\sqrt{x} = x^{1/2}$, so $4\sqrt{x} + 1 = 4x^{1/2} + 1$
- Numerator: $5x^{3/2} = 5x \cdot x^{1/2}$
- Denominator: $4x^{1/2} + 1$
- Divide numerator and denominator by $x^{1/2}$:
$$
\frac{5x}{4 + \frac{1}{x^{1/2}}} \to \frac{\infty}{4} = \infty
$$

✔ Answer: $\infty$

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Exercises 19–32: Find the limit

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

- Degrees equal (1), divide by $x$:
$$
\frac{2 - \frac{1}{x}}{3 + \frac{2}{x}} \to \frac{2}{3}
$$

✔ Answer: $\frac{2}{3}$

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#### 20. $\lim_{x \to \infty} \frac{3x^3 + 2}{9x^3 - 2x^2 + 7}$

- Degrees equal (3), divide by $x^3$:
$$
\frac{3 + \frac{2}{x^3}}{9 - \frac{2}{x} + \frac{7}{x^3}} \to \frac{3}{9} = \frac{1}{3}
$$

✔ Answer: $\frac{1}{3}$

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

- Degree denominator > numerator → limit is 0
- Divide by $x^2$:
$$
\frac{\frac{1}{x}}{1 - \frac{1}{x^2}} \to 0
$$

✔ Answer: $0$

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#### 22. $\lim_{x \to \infty} \left(4 + \frac{3}{x}\right)$

- As $x \to \infty$, $\frac{3}{x} \to 0$
- So, $4 + 0 = 4$

✔ Answer: $4$

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#### 23. $\lim_{x \to -\infty} \frac{5x^2}{x + 3}$

- Numerator degree: 2, denominator: 1 → numerator dominates
- As $x \to -\infty$, $5x^2 \to +\infty$, $x + 3 \to -\infty$
- So, $\frac{+\infty}{-\infty} = -\infty$

✔ Answer: $-\infty$

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#### 24. $\lim_{x \to -\infty} \left(\frac{1}{x} - \frac{4}{x^2}\right)$

- As $x \to -\infty$:
$\frac{1}{x} \to 0$, $\frac{4}{x^2} \to 0$
- So, $0 - 0 = 0$

✔ Answer: $0$

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#### 25. $\lim_{x \to -\infty} \frac{x}{\sqrt{x^2 + x}}$

- Note: $x < 0$ as $x \to -\infty$
- Factor out $x^2$ inside square root:
$\sqrt{x^2 + x} = |x|\sqrt{1 + \frac{1}{x}} = -x \sqrt{1 + \frac{1}{x}}$ because $x < 0$, so $|x| = -x$

So:
$$
\frac{x}{-x \sqrt{1 + \frac{1}{x}}} = \frac{1}{-\sqrt{1 + \frac{1}{x}}} \to \frac{1}{-\sqrt{1 + 0}} = -1
$$

✔ Answer: $-1$

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

- Similar to above: $x < 0$, $\sqrt{x^2 + 1} = |x|\sqrt{1 + \frac{1}{x^2}} = -x \sqrt{1 + \frac{1}{x^2}}$
- So:
$$
\frac{x}{-x \sqrt{1 + \frac{1}{x^2}}} = \frac{1}{-\sqrt{1 + 0}} = -1
$$

✔ Answer: $-1$

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

- Again, $x < 0$, so factor $x^2$ under square root:
$\sqrt{x^2 - x} = |x|\sqrt{1 - \frac{1}{x}} = -x \sqrt{1 - \frac{1}{x}}$

Numerator: $2x + 1 \approx 2x$

So:
$$
\frac{2x + 1}{-x \sqrt{1 - \frac{1}{x}}} = \frac{2 + \frac{1}{x}}{-\sqrt{1 - \frac{1}{x}}} \to \frac{2}{-\sqrt{1}} = -2
$$

✔ Answer: $-2$

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

- $x < 0$, so $\sqrt{x^2 + x} = |x|\sqrt{1 + \frac{1}{x}} = -x \sqrt{1 + \frac{1}{x}}$

Numerator: $-3x + 1 \approx -3x$, and since $x < 0$, $-3x > 0$

So:
$$
\frac{-3x + 1}{-x \sqrt{1 + \frac{1}{x}}} = \frac{-3 + \frac{1}{x}}{-\sqrt{1 + \frac{1}{x}}} \to \frac{-3}{-1} = 3
$$

✔ Answer: $3$

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#### 29. $\lim_{x \to \infty} \frac{\sin 2x}{x}$

- $|\sin 2x| \leq 1$, so $|\frac{\sin 2x}{x}| \leq \frac{1}{x} \to 0$
- By Squeeze Theorem: limit is 0

✔ Answer: $0$

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#### 30. $\lim_{x \to \infty} \frac{x - \cos x}{x}$

- Split:
$$
\frac{x}{x} - \frac{\cos x}{x} = 1 - \frac{\cos x}{x}
$$
- $|\cos x| \leq 1$, so $\frac{\cos x}{x} \to 0$
- So limit is $1 - 0 = 1$

✔ Answer: $1$

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#### 31. $\lim_{x \to \infty} \frac{1}{2x + \sin x}$

- $|\sin x| \leq 1$, so $2x + \sin x \geq 2x - 1$ → as $x \to \infty$, this goes to $\infty$
- So denominator → $\infty$, numerator = 1 → limit is 0

✔ Answer: $0$

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#### 32. $\lim_{x \to \infty} \cos \frac{1}{x}$

- As $x \to \infty$, $\frac{1}{x} \to 0$
- So $\cos \frac{1}{x} \to \cos 0 = 1$

✔ Answer: $1$

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 15(a) | $-2$ |
| 15(b) | $1$ |
| 15(c) | $\infty$ |
| 16(a) | $0$ |
| 16(b) | $-\frac{2}{3}$ |
| 16(c) | $-\infty$ |
| 17(a) | $0$ |
| 17(b) | $-\frac{2}{3}$ |
| 17(c) | $-\infty$ |
| 18(a) | $0$ |
| 18(b) | $\frac{5}{4}$ |
| 18(c) | $\infty$ |
| 19 | $\frac{2}{3}$ |
| 20 | $\frac{1}{3}$ |
| 21 | $0$ |
| 22 | $4$ |
| 23 | $-\infty$ |
| 24 | $0$ |
| 25 | $-1$ |
| 26 | $-1$ |
| 27 | $-2$ |
| 28 | $3$ |
| 29 | $0$ |
| 30 | $1$ |
| 31 | $0$ |
| 32 | $1$ |

Let me know if you'd like a visual explanation or graph for any of these!