Trigonometric limits worksheet for calculus practice.

Let's solve each of these limit problems step by step. These are all limits involving trigonometric functions as $ x \to 0 $. We will use the standard limit:

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

Also recall:
- $ \sin(ax) \approx ax $ as $ x \to 0 $
- $ \tan(ax) \approx ax $ as $ x \to 0 $

We'll use these approximations or algebraic manipulation to evaluate each limit.

---

1. $ \lim_{x \to 0} \frac{\sin x}{3x} $

$$
= \frac{1}{3} \cdot \frac{\sin x}{x} \to \frac{1}{3} \cdot 1 = \boxed{\frac{1}{3}}
$$

---

2. $ \lim_{x \to 0} \frac{\sin x}{\frac{1}{3}x} $

$$
= \frac{\sin x}{x} \cdot \frac{1}{1/3} = 1 \cdot 3 = \boxed{3}
$$

---

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

Let $ u = 3x $, so as $ x \to 0 $, $ u \to 0 $. Then:

$$
= \frac{\sin u}{u/3} = 3 \cdot \frac{\sin u}{u} \to 3 \cdot 1 = \boxed{3}
$$

---

4. $ \lim_{x \to 0} \frac{\sin 6x}{2x} $

$$
= \frac{6}{2} \cdot \frac{\sin 6x}{6x} = 3 \cdot 1 = \boxed{3}
$$

---

5. $ \lim_{x \to 0} \frac{\sin x}{3x} $

Same as problem 1:
$$
= \frac{1}{3} \cdot \frac{\sin x}{x} \to \frac{1}{3} \cdot 1 = \boxed{\frac{1}{3}}
$$

---

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

$$
= \frac{3x}{\sin 3x} = \frac{1}{\frac{\sin 3x}{3x}} \to \frac{1}{1} = \boxed{1}
$$

---

7. $ \lim_{x \to 0} \frac{\tan 2x}{4x} $

Use $ \tan 2x \sim 2x $ as $ x \to 0 $:

$$
= \frac{2x}{4x} = \frac{2}{4} = \boxed{\frac{1}{2}}
$$

---

8. $ \lim_{x \to 0} \frac{\tan 2x}{\sin 4x} $

Use approximations:
- $ \tan 2x \sim 2x $
- $ \sin 4x \sim 4x $

So:
$$
\frac{2x}{4x} = \frac{2}{4} = \boxed{\frac{1}{2}}
$$

Alternatively, write:
$$
= \frac{\tan 2x}{2x} \cdot \frac{2x}{\sin 4x} = \left( \frac{\tan 2x}{2x} \right) \cdot \left( \frac{2x}{4x} \cdot \frac{4x}{\sin 4x} \right)
= (1) \cdot \left( \frac{1}{2} \cdot 1 \right) = \frac{1}{2}
$$

---

9. $ \lim_{x \to 0} \frac{6x}{\tan \frac{1}{2}x} $

Note: $ \tan \left(\frac{1}{2}x\right) \sim \frac{1}{2}x $, so:

$$
= \frac{6x}{\frac{1}{2}x} = \frac{6}{1/2} = 12 \Rightarrow \boxed{12}
$$

---

10. $ \lim_{x \to 0} \frac{\sin 3x}{\tan 2x} $

Use approximations:
- $ \sin 3x \sim 3x $
- $ \tan 2x \sim 2x $

So:
$$
\frac{3x}{2x} = \frac{3}{2} \Rightarrow \boxed{\frac{3}{2}}
$$

---

11. $ \lim_{x \to 0} \frac{\sin ax}{\sin bx} $

As $ x \to 0 $, $ \sin ax \sim ax $, $ \sin bx \sim bx $, so:

$$
\frac{ax}{bx} = \frac{a}{b} \Rightarrow \boxed{\frac{a}{b}}
$$

---

12. $ \lim_{x \to 0} \frac{\tan ax}{\tan bx} $

Similarly:
- $ \tan ax \sim ax $
- $ \tan bx \sim bx $

So:
$$
\frac{ax}{bx} = \frac{a}{b} \Rightarrow \boxed{\frac{a}{b}}
$$

---

13. $ \lim_{x \to 0} \frac{\sin^2 6x}{2x} $

Note: $ \sin^2 6x = (\sin 6x)^2 \sim (6x)^2 = 36x^2 $

So:
$$
\frac{36x^2}{2x} = 18x \to 0 \quad \text{as } x \to 0
\Rightarrow \boxed{0}
$$

---

14. $ \lim_{x \to 0} \frac{\sin^2 x}{3x} $

$ \sin^2 x \sim x^2 $, so:

$$
\frac{x^2}{3x} = \frac{x}{3} \to 0 \Rightarrow \boxed{0}
$$

---

15. $ \lim_{x \to 0} \frac{3x}{\sin^2 3x} $

$ \sin^2 3x \sim (3x)^2 = 9x^2 $

So:
$$
\frac{3x}{9x^2} = \frac{1}{3x} \to \infty \quad \text{(diverges)}
$$

But let’s be careful: numerator → 0, denominator → 0, but faster.

As $ x \to 0 $, $ \frac{3x}{9x^2} = \frac{1}{3x} \to \infty $ if $ x \to 0^+ $, and $ -\infty $ if $ x \to 0^- $. So the limit does not exist in finite sense.

But since it's a two-sided limit and behavior differs, we say:

$$
\boxed{\text{does not exist}} \quad \text{or} \quad \boxed{\infty} \text{ (if one-sided is intended)}
$$

But usually in such contexts, they expect:

Wait — actually, check again:

$ \sin^2 3x \sim 9x^2 $, so:

$$
\frac{3x}{\sin^2 3x} \sim \frac{3x}{9x^2} = \frac{1}{3x} \to \pm\infty
$$

So limit does not exist (infinite divergence).

But sometimes written as $ \infty $ or undefined.

But likely expected answer: $ \boxed{\infty} $ or DNE

However, perhaps typo? Let’s see.

Wait — maybe it should be $ \frac{3x^2}{\sin^2 3x} $? But no, as written.

So final answer: $ \boxed{\text{does not exist}} $ (or $ \infty $ depending on convention)

But let’s keep it as: $ \boxed{\infty} $ if we consider magnitude, but strictly speaking, no finite limit.

But for consistency with others, let's write:

$$
\boxed{\infty} \quad \text{(since magnitude goes to infinity)}
$$

But better: limit does not exist due to infinite oscillation.

But in many textbooks, they write $ \infty $ when it diverges to infinity.

So I’ll go with: $ \boxed{\infty} $, though technically it's not defined.

But wait — $ \frac{3x}{\sin^2 3x} \sim \frac{3x}{9x^2} = \frac{1}{3x} $, which tends to $ +\infty $ as $ x \to 0^+ $, $ -\infty $ as $ x \to 0^- $. So two-sided limit does not exist.

But if only $ x \to 0^+ $, then $ \infty $.

Assuming two-sided, answer is: $ \boxed{\text{DNE}} $

But perhaps the problem assumes $ x \to 0^+ $. Since most limits like this assume positive approach.

But let’s move on.

---

16. $ \lim_{x \to 0} \frac{\tan^2 ab}{\tan^2 bx} $

Wait — here's a problem: "ab" is constant?

Look: $ \tan^2 ab $ — is $ ab $ a constant?

But $ x \to 0 $, and $ ab $ is not depending on $ x $? That would mean $ \tan^2(ab) $ is constant, while denominator $ \tan^2(bx) \to 0 $.

So unless $ ab $ is meant to be $ a x $ or $ a b x $, this doesn't make sense.

Likely typo: probably meant $ \tan^2(ax) $ or $ \tan^2(a x) $, but written as $ \tan^2 ab $.

But "ab" is likely $ a \cdot b $, constants.

Then $ \tan^2(ab) $ is a constant (say $ C $), and $ \tan^2(bx) \to 0 $ as $ x \to 0 $.

So:
$$
\frac{C}{\tan^2(bx)} \to \infty \quad \text{if } C \ne 0
$$

But $ \tan^2(ab) $ is fixed, not zero unless $ ab = n\pi $.

So assuming $ ab \ne n\pi $, then $ \tan^2(ab) \ne 0 $, so:

$$
\frac{\text{constant}}{0^+} \to \infty
\Rightarrow \boxed{\infty}
$$

But this seems odd. Probably intended to be:

Problem 16: $ \lim_{x \to 0} \frac{\tan^2(ax)}{\tan^2(bx)} $

That makes more sense.

Similarly for 17 and 18.

Let me assume that “ab” is a typo and should be “ax” or “a x”.

But looking at 17: $ \lim_{x \to 0} \frac{\tan ab}{\tan^2 bx} $ — same issue.

So likely, “ab” should be “ax”, i.e., $ \tan(ax) $, etc.

Let me correct based on context.

Probably:
- 16: $ \lim_{x \to 0} \frac{\tan^2(ax)}{\tan^2(bx)} $
- 17: $ \lim_{x \to 0} \frac{\tan(ax)}{\tan^2(bx)} $
- 18: $ \lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} $

That makes sense.

I'll assume that.

---

16. $ \lim_{x \to 0} \frac{\tan^2(ax)}{\tan^2(bx)} $

Use $ \tan(ax) \sim ax $, $ \tan(bx) \sim bx $:

$$
\frac{(ax)^2}{(bx)^2} = \frac{a^2 x^2}{b^2 x^2} = \frac{a^2}{b^2} \Rightarrow \boxed{\frac{a^2}{b^2}}
$$

---

17. $ \lim_{x \to 0} \frac{\tan(ax)}{\tan^2(bx)} $

$ \tan(ax) \sim ax $, $ \tan^2(bx) \sim (bx)^2 = b^2 x^2 $

So:
$$
\frac{ax}{b^2 x^2} = \frac{a}{b^2 x} \to \infty \quad \text{as } x \to 0
$$

Again, sign depends on side. So limit does not exist (infinite).

But magnitude $ \to \infty $, so often written as $ \boxed{\infty} $, but strictly DNE.

But if $ a = 0 $, it would be 0, but assume $ a \ne 0 $.

So: $ \boxed{\infty} $ or $ \boxed{\text{DNE}} $

But given pattern, likely: $ \boxed{\infty} $

But let’s see.

Wait: numerator ~ $ ax $, denominator ~ $ b^2 x^2 $, so ratio ~ $ \frac{a}{b^2 x} \to \pm\infty $

So two-sided limit does not exist.

But in many contexts, they accept $ \infty $.

But let’s write: $ \boxed{\infty} $ (assuming $ x \to 0^+ $)

But to be precise, limit does not exist.

But for now, I’ll write $ \boxed{\infty} $

---

18. $ \lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} $

$ \tan(ax) \sim ax $, $ \tan(bx) \sim bx $

So:
$$
\frac{ax}{bx} = \frac{a}{b} \Rightarrow \boxed{\frac{a}{b}}
$$

---

Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ \frac{1}{3} $ |
| 2 | $ 3 $ |
| 3 | $ 3 $ |
| 4 | $ 3 $ |
| 5 | $ \frac{1}{3} $ |
| 6 | $ 1 $ |
| 7 | $ \frac{1}{2} $ |
| 8 | $ \frac{1}{2} $ |
| 9 | $ 12 $ |
| 10 | $ \frac{3}{2} $ |
| 11 | $ \frac{a}{b} $ |
| 12 | $ \frac{a}{b} $ |
| 13 | $ 0 $ |
| 14 | $ 0 $ |
| 15 | $ \infty $ (or DNE) |
| 16 | $ \frac{a^2}{b^2} $ |
| 17 | $ \infty $ (or DNE) |
| 18 | $ \frac{a}{b} $ |

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Notes:

- Problems 15 and 17 involve division by a quantity going to zero faster than numerator, leading to infinite limits.
- For 15: $ \frac{3x}{\sin^2 3x} \sim \frac{3x}{9x^2} = \frac{1}{3x} \to \infty $
- For 17: $ \frac{\tan(ax)}{\tan^2(bx)} \sim \frac{ax}{b^2 x^2} = \frac{a}{b^2 x} \to \infty $
- Assumed "ab" in 16–18 means $ ax $, $ bx $, not constant product.

If "ab" is really a constant, then:
- 16: $ \frac{\tan^2(ab)}{\tan^2(bx)} \to \infty $ (since denominator → 0, numerator constant ≠ 0)
- 17: $ \frac{\tan(ab)}{\tan^2(bx)} \to \infty $
- 18: $ \frac{\tan(ab)}{\tan(bx)} \to \infty $ (since denominator → 0, numerator constant)

But that would make answers always $ \infty $, which seems less meaningful.

Hence, most likely typo, and should be $ \tan(ax) $, $ \tan(bx) $.

So final corrected answers:

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✔ Corrected Final Answers:

1. $ \boxed{\dfrac{1}{3}} $
2. $ \boxed{3} $
3. $ \boxed{3} $
4. $ \boxed{3} $
5. $ \boxed{\dfrac{1}{3}} $
6. $ \boxed{1} $
7. $ \boxed{\dfrac{1}{2}} $
8. $ \boxed{\dfrac{1}{2}} $
9. $ \boxed{12} $
10. $ \boxed{\dfrac{3}{2}} $
11. $ \boxed{\dfrac{a}{b}} $
12. $ \boxed{\dfrac{a}{b}} $
13. $ \boxed{0} $
14. $ \boxed{0} $
15. $ \boxed{\infty} $ (limit does not exist)
16. $ \boxed{\dfrac{a^2}{b^2}} $
17. $ \boxed{\infty} $ (limit does not exist)
18. $ \boxed{\dfrac{a}{b}} $

> Note: For 15 and 17, the limit is infinite; it does not exist in the finite sense. If required, specify "does not exist" or "tends to infinity".