Trigonometric limits worksheet for math tutorial.

Let's solve each of these limits step by step, focusing on limits of trigonometric functions, especially using standard limit results like:

$$
\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0, \quad \lim_{x \to 0} \frac{\tan x}{x} = 1
$$

We’ll go through the problems one by one.

---

1. $\displaystyle \lim_{\theta \to 0} \frac{0}{\cos \theta}$

- Numerator is $0$, denominator approaches $\cos(0) = 1$.
- So: $\frac{0}{1} = 0$

✔ Answer: $0$

---

2. $\displaystyle \lim_{x \to \infty} \sec\left(\frac{3}{x^2}\right)$

- As $x \to \infty$, $\frac{3}{x^2} \to 0$
- So $\sec\left(\frac{3}{x^2}\right) \to \sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1$

✔ Answer: $1$

---

3. $\displaystyle \lim_{\theta \to 0} \frac{\sin 3\theta}{\theta}$

- Use identity: $\frac{\sin 3\theta}{\theta} = 3 \cdot \frac{\sin 3\theta}{3\theta}$
- As $\theta \to 0$, $3\theta \to 0$, so $\frac{\sin 3\theta}{3\theta} \to 1$
- So: $3 \cdot 1 = 3$

✔ Answer: $3$

---

4. $\displaystyle \lim_{x \to 0} \frac{\sin 2x}{3x}$

- Rewrite: $\frac{\sin 2x}{3x} = \frac{2}{3} \cdot \frac{\sin 2x}{2x}$
- As $x \to 0$, $\frac{\sin 2x}{2x} \to 1$
- So: $\frac{2}{3} \cdot 1 = \frac{2}{3}$

✔ Answer: $\frac{2}{3}$

---

5. $\displaystyle \lim_{k \to 0} \frac{\sin\left(\frac{k}{2}\right)}{k}$

- Write: $\frac{\sin(k/2)}{k} = \frac{1}{2} \cdot \frac{\sin(k/2)}{k/2}$
- As $k \to 0$, $k/2 \to 0$, so $\frac{\sin(k/2)}{k/2} \to 1$
- So: $\frac{1}{2} \cdot 1 = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

---

6. $\displaystyle \lim_{x \to 0^+} \frac{\sin x}{\sqrt{x}}$

- As $x \to 0^+$, $\sin x \sim x$, so:
$$
\frac{\sin x}{\sqrt{x}} \sim \frac{x}{\sqrt{x}} = \sqrt{x} \to 0
$$
- Since $\sin x > 0$ for small $x > 0$, and $\sqrt{x} \to 0^+$, the expression goes to $0$

✔ Answer: $0$

---

7. $\displaystyle \lim_{x \to 0} \frac{\sin^2 x}{x}$

- $\sin^2 x = (\sin x)^2$, so:
$$
\frac{\sin^2 x}{x} = \frac{\sin x}{x} \cdot \sin x
$$
- As $x \to 0$: $\frac{\sin x}{x} \to 1$, $\sin x \to 0$
- So product: $1 \cdot 0 = 0$

✔ Answer: $0$

---

8. $\displaystyle \lim_{y \to 0} \frac{\sin y}{y(y + 3)}$

- Split: $\frac{\sin y}{y} \cdot \frac{1}{y + 3}$
- As $y \to 0$: $\frac{\sin y}{y} \to 1$, $\frac{1}{y+3} \to \frac{1}{3}$
- So: $1 \cdot \frac{1}{3} = \frac{1}{3}$

✔ Answer: $\frac{1}{3}$

---

9. $\displaystyle \lim_{x \to 1} \frac{\sin(x - 1)}{x^2 - 1}$

- Note: $x^2 - 1 = (x - 1)(x + 1)$
- Let $u = x - 1$, then as $x \to 1$, $u \to 0$
- So:
$$
\frac{\sin u}{(u + 1)(u + 2)} = \frac{\sin u}{u} \cdot \frac{1}{(u + 1)(u + 2)}
$$
- As $u \to 0$: $\frac{\sin u}{u} \to 1$, denominator $\to (1)(2) = 2$
- So limit: $1 \cdot \frac{1}{2} = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

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10. $\displaystyle \lim_{\theta \to 0} \frac{\sin \theta \cos \theta}{\theta}$

- Use identity: $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$
- So:
$$
\frac{\sin \theta \cos \theta}{\theta} = \frac{1}{2} \cdot \frac{\sin 2\theta}{\theta} = \frac{1}{2} \cdot 2 \cdot \frac{\sin 2\theta}{2\theta} = \frac{\sin 2\theta}{2\theta} \to 1
$$
- So overall: $1$

Alternatively:
$$
\frac{\sin \theta \cos \theta}{\theta} = \frac{\sin \theta}{\theta} \cdot \cos \theta \to 1 \cdot 1 = 1
$$

✔ Answer: $1$

---

11. $\displaystyle \lim_{x \to 0} \frac{\sin 2x}{x \cos x}$

- $\frac{\sin 2x}{x \cos x} = \frac{2 \sin x \cos x}{x \cos x} = \frac{2 \sin x}{x}$
- As $x \to 0$: $\frac{\sin x}{x} \to 1$, so $2 \cdot 1 = 2$

✔ Answer: $2$

---

12. $\displaystyle \lim_{x \to 0} \frac{\sin 3x}{x \cos x}$

- $\frac{\sin 3x}{x \cos x} = \frac{\sin 3x}{3x} \cdot \frac{3}{\cos x}$
- As $x \to 0$: $\frac{\sin 3x}{3x} \to 1$, $\cos x \to 1$, so:
$$
1 \cdot \frac{3}{1} = 3
$$

✔ Answer: $3$

---

13. $\displaystyle \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{x \cos x}$

- $\frac{\sin(x/2)}{x \cos x} = \frac{\sin(x/2)}{x/2} \cdot \frac{1}{2 \cos x}$
- As $x \to 0$: $\frac{\sin(x/2)}{x/2} \to 1$, $\cos x \to 1$
- So: $1 \cdot \frac{1}{2} = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

---

14. $\displaystyle \lim_{\theta \to 0} \frac{2 \tan \theta}{\theta}$

- $\tan \theta = \frac{\sin \theta}{\cos \theta}$, so:
$$
\frac{2 \tan \theta}{\theta} = 2 \cdot \frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta}
$$
- As $\theta \to 0$: $\frac{\sin \theta}{\theta} \to 1$, $\cos \theta \to 1$
- So: $2 \cdot 1 \cdot 1 = 2$

✔ Answer: $2$

---

15. $\displaystyle \lim_{x \to 0} \frac{x}{\tan x}$

- $\frac{x}{\tan x} = \frac{x}{\sin x / \cos x} = \frac{x \cos x}{\sin x} = \frac{x}{\sin x} \cdot \cos x$
- As $x \to 0$: $\frac{x}{\sin x} \to 1$, $\cos x \to 1$
- So: $1 \cdot 1 = 1$

✔ Answer: $1$

---

16. $\displaystyle \lim_{x \to 0} \frac{x}{\tan x}$

Same as #15 → Answer: 1

Wait — this appears twice? Actually, #15 and #16 are the same.

But looking back: #15 was not listed in your image — perhaps typo.

In your list:

> 16. $\displaystyle \lim_{x \to 0} \frac{x}{\tan x}$

Yes, that’s correct. So it's same as #15.

✔ Answer: $1$

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17. $\displaystyle \lim_{x \to 0} \frac{\tan 6x}{\sin 3x}$

- $\tan 6x = \frac{\sin 6x}{\cos 6x}$, so:
$$
\frac{\tan 6x}{\sin 3x} = \frac{\sin 6x}{\cos 6x \cdot \sin 3x}
$$
- Now write $\sin 6x = 2 \sin 3x \cos 3x$, so:
$$
\frac{2 \sin 3x \cos 3x}{\cos 6x \cdot \sin 3x} = \frac{2 \cos 3x}{\cos 6x}
$$
- As $x \to 0$: $\cos 3x \to 1$, $\cos 6x \to 1$, so:
$$
\frac{2 \cdot 1}{1} = 2
$$

✔ Answer: $2$

---

18. $\displaystyle \lim_{x \to 0} \frac{\sin 6x}{\tan 5x}$

- $\tan 5x = \frac{\sin 5x}{\cos 5x}$, so:
$$
\frac{\sin 6x}{\tan 5x} = \frac{\sin 6x \cdot \cos 5x}{\sin 5x}
$$
- Now split:
$$
= \frac{\sin 6x}{6x} \cdot \frac{6x}{5x} \cdot \frac{\sin 5x}{5x}^{-1} \cdot \cos 5x
$$
Better:
$$
= \frac{\sin 6x}{6x} \cdot \frac{6}{5} \cdot \frac{5x}{\sin 5x} \cdot \cos 5x
$$
As $x \to 0$:
- $\frac{\sin 6x}{6x} \to 1$
- $\frac{5x}{\sin 5x} \to 1$
- $\cos 5x \to 1$
So:
$$
1 \cdot \frac{6}{5} \cdot 1 \cdot 1 = \frac{6}{5}
$$

✔ Answer: $\frac{6}{5}$

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19. $\displaystyle \lim_{x \to 0} \frac{x(x + 2)}{\sin x}$

- $\frac{x(x+2)}{\sin x} = \frac{x}{\sin x} \cdot (x + 2)$
- As $x \to 0$: $\frac{x}{\sin x} \to 1$, $x + 2 \to 2$
- So: $1 \cdot 2 = 2$

✔ Answer: $2$

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20. $\displaystyle \lim_{x \to 0} \frac{\sin 5x}{2x(1 + 4\cos 2x)}$

- $\frac{\sin 5x}{2x(1 + 4\cos 2x)} = \frac{5}{2} \cdot \frac{\sin 5x}{5x} \cdot \frac{1}{1 + 4\cos 2x}$
- As $x \to 0$: $\frac{\sin 5x}{5x} \to 1$, $\cos 2x \to 1$, so denominator: $1 + 4(1) = 5$
- So: $\frac{5}{2} \cdot 1 \cdot \frac{1}{5} = \frac{1}{2}$

✔ Answer: $\frac{1}{2}$

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21. $\displaystyle \lim_{x \to 0} \frac{\sin 4x}{2x(1 + \cos 3x)}$

- $\frac{\sin 4x}{2x(1 + \cos 3x)} = \frac{4}{2} \cdot \frac{\sin 4x}{4x} \cdot \frac{1}{1 + \cos 3x} = 2 \cdot \frac{\sin 4x}{4x} \cdot \frac{1}{1 + \cos 3x}$
- As $x \to 0$: $\frac{\sin 4x}{4x} \to 1$, $\cos 3x \to 1$, so $1 + \cos 3x \to 2$
- So: $2 \cdot 1 \cdot \frac{1}{2} = 1$

✔ Answer: $1$

---

22. $\displaystyle \lim_{x \to 0} \frac{x^2 - 3\sin x}{x}$

- Split: $\frac{x^2}{x} - \frac{3\sin x}{x} = x - 3 \cdot \frac{\sin x}{x}$
- As $x \to 0$: $x \to 0$, $\frac{\sin x}{x} \to 1$
- So: $0 - 3(1) = -3$

✔ Answer: $-3$

---

23. $\displaystyle \lim_{x \to 0} \frac{\sin x + 2x}{x}$

- $\frac{\sin x}{x} + \frac{2x}{x} = \frac{\sin x}{x} + 2 \to 1 + 2 = 3$

✔ Answer: $3$

---

24. $\displaystyle \lim_{x \to 0} \frac{5x - \sin 5x}{2x}$

- $\frac{5x - \sin 5x}{2x} = \frac{5x}{2x} - \frac{\sin 5x}{2x} = \frac{5}{2} - \frac{5}{2} \cdot \frac{\sin 5x}{5x}$
- As $x \to 0$: $\frac{\sin 5x}{5x} \to 1$, so:
$$
\frac{5}{2} - \frac{5}{2} \cdot 1 = 0
$$

✔ Answer: $0$

---

25. $\displaystyle \lim_{x \to 0} \frac{x^2 + 3\sin 5x}{x}$

- $\frac{x^2}{x} + \frac{3\sin 5x}{x} = x + 3 \cdot \frac{\sin 5x}{x} = x + 3 \cdot 5 \cdot \frac{\sin 5x}{5x} = x + 15 \cdot \frac{\sin 5x}{5x}$
- As $x \to 0$: $x \to 0$, $\frac{\sin 5x}{5x} \to 1$, so total: $0 + 15 = 15$

✔ Answer: $15$

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26. $\displaystyle \lim_{x \to 0} \frac{x^2 + 4\sin 7x}{x}$

- Similar: $\frac{x^2}{x} + \frac{4\sin 7x}{x} = x + 4 \cdot 7 \cdot \frac{\sin 7x}{7x} = x + 28 \cdot \frac{\sin 7x}{7x}$
- As $x \to 0$: $x \to 0$, $\frac{\sin 7x}{7x} \to 1$, so total: $0 + 28 = 28$

✔ Answer: $28$

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27. $\displaystyle \lim_{x \to 0} \frac{x^2 + 4\sin 3x}{x}$

- Same method: $\frac{x^2}{x} + \frac{4\sin 3x}{x} = x + 4 \cdot 3 \cdot \frac{\sin 3x}{3x} = x + 12 \cdot \frac{\sin 3x}{3x}$
- As $x \to 0$: $x \to 0$, ratio $\to 1$, so answer: $12$

✔ Answer: $12$

---

28. $\displaystyle \lim_{x \to 0} \frac{x \cos x - 3\sin x}{x}$

- Split: $\frac{x \cos x}{x} - \frac{3\sin x}{x} = \cos x - 3 \cdot \frac{\sin x}{x}$
- As $x \to 0$: $\cos x \to 1$, $\frac{\sin x}{x} \to 1$
- So: $1 - 3(1) = -2$

✔ Answer: $-2$

---

29. $\displaystyle \lim_{x \to 0} \frac{x \sin 2x - \sin 3x}{6x}$

- Break into two terms:
$$
\frac{x \sin 2x}{6x} - \frac{\sin 3x}{6x} = \frac{\sin 2x}{6} - \frac{\sin 3x}{6x}
$$
- First term: $\frac{\sin 2x}{6} \to \frac{0}{6} = 0$ as $x \to 0$
- Second term: $\frac{\sin 3x}{6x} = \frac{3}{6} \cdot \frac{\sin 3x}{3x} = \frac{1}{2} \cdot 1 = \frac{1}{2}$, but with a minus sign
- So:
$$
\frac{\sin 2x}{6} - \frac{\sin 3x}{6x} \to 0 - \frac{1}{2} = -\frac{1}{2}
$$

Wait — let's be careful:

Actually:
$$
\frac{x \sin 2x - \sin 3x}{6x} = \frac{x \sin 2x}{6x} - \frac{\sin 3x}{6x} = \frac{\sin 2x}{6} - \frac{\sin 3x}{6x}
$$
Now:
- $\frac{\sin 2x}{6} \to 0$
- $\frac{\sin 3x}{6x} = \frac{3}{6} \cdot \frac{\sin 3x}{3x} = \frac{1}{2} \cdot 1 = \frac{1}{2}$
- So: $0 - \frac{1}{2} = -\frac{1}{2}$

✔ Answer: $-\frac{1}{2}$

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## ✔ Final Answers Summary:

| No | Answer |
|----|--------|
| 1 | $0$ |
| 2 | $1$ |
| 3 | $3$ |
| 4 | $\frac{2}{3}$ |
| 5 | $\frac{1}{2}$ |
| 6 | $0$ |
| 7 | $0$ |
| 8 | $\frac{1}{3}$ |
| 9 | $\frac{1}{2}$ |
| 10 | $1$ |
| 11 | $2$ |
| 12 | $3$ |
| 13 | $\frac{1}{2}$ |
| 14 | $2$ |
| 15 | $1$ |
| 16 | $1$ |
| 17 | $2$ |
| 18 | $\frac{6}{5}$ |
| 19 | $2$ |
| 20 | $\frac{1}{2}$ |
| 21 | $1$ |
| 22 | $-3$ |
| 23 | $3$ |
| 24 | $0$ |
| 25 | $15$ |
| 26 | $28$ |
| 27 | $12$ |
| 28 | $-2$ |
| 29 | $-\frac{1}{2}$ |

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