Limit of Trigonometric Functions worksheet - Free Printable

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:
- $ \tan x = \frac{\sin x}{\cos x} $
- For small $ x $, $ \sin x \approx x $, $ \tan x \approx x $

We’ll apply algebraic manipulation and substitution where needed.

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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} $

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

---

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

Let $ u = 3x $, then as $ x \to 0 $, $ u \to 0 $:

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

---

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 #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 $:

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

Or formally:

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

---

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

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

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

Or more rigorously:

$$
= \frac{\tan 2x}{2x} \cdot \frac{2x}{\sin 4x} = 1 \cdot \frac{2x}{4x} = \frac{1}{2} \cdot \frac{\sin 4x}{4x}^{-1} \cdot \frac{2x}{\sin 4x} \text{ wait — better way:}

\frac{\tan 2x}{\sin 4x} = \frac{\tan 2x}{2x} \cdot \frac{2x}{\sin 4x} = 1 \cdot \frac{2x}{4x} \cdot \frac{4x}{\sin 4x} = \frac{1}{2} \cdot 1 = \boxed{\frac{1}{2}}
$$

---

9. $ \lim_{x \to 0} \frac{6x}{\tan \left(\frac{1}{2}x\right)} $

As $ x \to 0 $, $ \tan\left(\frac{1}{2}x\right) \sim \frac{1}{2}x $

So:

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

More formally:

$$
= \frac{6x}{\frac{1}{2}x} \cdot \frac{\frac{1}{2}x}{\tan\left(\frac{1}{2}x\right)} = 12 \cdot 1 = \boxed{12}
$$

---

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

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

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

Formally:

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

---

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

Use $ \sin ax \sim ax $, $ \sin bx \sim bx $

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

---

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

$ \tan ax \sim ax $, $ \tan bx \sim bx $

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

---

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

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

So:

$$
\frac{36x^2}{2x} = 18x \to 0 \Rightarrow \boxed{0}
$$

---

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

Then:

$$
\frac{3x}{9x^2} = \frac{1}{3x} \to \infty \Rightarrow \boxed{\infty}
$$

But wait: as $ x \to 0^+ $, it goes to $ +\infty $; $ x \to 0^- $, $ -\infty $. So limit does not exist in real numbers?

Wait: numerator $ 3x \to 0 $, denominator $ \sin^2 3x \to 0^+ $, always positive.

So $ \frac{3x}{\sin^2 3x} \to 0 $ if $ x \to 0 $, because numerator goes to zero faster than denominator? Wait:

Numerator: $ O(x) $, denominator: $ O(x^2) $ → ratio $ \sim \frac{x}{x^2} = \frac{1}{x} \to \pm\infty $

So:
- As $ x \to 0^+ $: $ \frac{3x}{\sin^2 3x} \to +\infty $
- As $ x \to 0^- $: $ \to -\infty $

So the two-sided limit does not exist.

But often in such contexts, they may expect magnitude or consider absolute value.

But strictly speaking:

$$
\lim_{x \to 0} \frac{3x}{\sin^2 3x} = \text{DNE (does not exist)}
$$

But let’s check:

Let $ u = 3x $, so $ x = u/3 $, as $ x \to 0 $, $ u \to 0 $

$$
= \frac{3(u/3)}{\sin^2 u} = \frac{u}{\sin^2 u} = \frac{1}{\sin u} \cdot \frac{u}{\sin u} \to \frac{1}{0} \cdot 1 = \infty \text{ or } -\infty
$$

But since $ \sin u \to 0 $, and sign depends on side.

So limit does not exist.

But perhaps the problem assumes $ x \to 0^+ $? Or maybe typo?

Alternatively, maybe the expression is meant to be $ \frac{3x}{\sin 3x} $, but no — it's $ \sin^2 3x $.

So answer: Does not exist, or infinite depending on convention.

But since $ \sin^2 3x > 0 $, and $ 3x $ changes sign, the limit does not exist.

But in many textbooks, if they write this, they might expect $ \boxed{\infty} $ assuming $ x \to 0^+ $, but strictly:

> $\boxed{\text{DNE}}$

However, looking at other problems, likely they expect magnitude, so maybe we write:

$$
\lim_{x \to 0} \frac{3x}{\sin^2 3x} = \boxed{\text{undefined (DNE)}}
$$

But let's see if it can be written as:

$$
= \frac{3x}{(3x)^2} \cdot \left(\frac{3x}{\sin 3x}\right)^2 = \frac{1}{3x} \cdot 1^2 \to \pm\infty
$$

So again, DNE.

But perhaps the problem has a typo? Let's skip for now.

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

Wait — ab? Is that a constant or $ a \cdot b $? But $ x \to 0 $, and $ ab $ is constant?

Wait — probably typo. Likely should be $ \tan^2(ax) $ over $ \tan^2(bx) $, or something.

But here it says $ \tan^2 ab $ — that would be $ \tan^2(ab) $, which is constant, not depending on $ x $.

So:

$$
\lim_{x \to 0} \frac{\tan^2(ab)}{\tan^2(bx)} = \frac{\tan^2(ab)}{0} \to \infty \quad \text{(if } \tan(ab) \ne 0\text{)}
$$

But $ \tan(ab) $ is constant, and $ \tan(bx) \to 0 $, so unless $ \tan(ab) = 0 $, this blows up.

But $ ab $ is likely meant to be $ ax $, not $ ab $.

Likely typo: probably meant:

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

Assuming that:

Then $ \tan(ax) \sim ax $, $ \tan(bx) \sim bx $, so:

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

Similarly, 17 and 18 likely have typos.

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

Again, $ \tan ab $ is constant, $ \tan^2 bx \to 0 $, so unless $ \tan ab = 0 $, this is infinite.

But likely meant:

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

Then $ \sim \frac{ax}{(bx)^2} = \frac{a}{b^2 x} \to \pm\infty $ → DNE

But if it's $ \frac{\tan(ax)}{\tan(bx)} $, then $ \sim \frac{ax}{bx} = \frac{a}{b} $

But as written: $ \frac{\tan ab}{\tan^2 bx} $ — doesn't make sense.

---

18. $ \lim_{x \to 0} \frac{\tan ab}{\tan bx} $

Same issue: $ \tan ab $ constant, $ \tan bx \to 0 $, so unless $ \tan ab = 0 $, this goes to $ \pm\infty $

So likely, all of 16–18 have typos.

Probably intended:

- 16: $ \lim_{x \to 0} \frac{\tan^2(ax)}{\tan^2(bx)} = \boxed{\left(\frac{a}{b}\right)^2} $
- 17: $ \lim_{x \to 0} \frac{\tan(ax)}{\tan^2(bx)} = \boxed{\infty} $ or DNE
- 18: $ \lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} = \boxed{\frac{a}{b}} $

But based on what's written, we must assume the variables are constants.

But $ ab $ is constant, $ bx $ depends on $ x $.

So let's re-express with correct interpretation.

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## Final Answers (assuming typos corrected):

| 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 | DNE (or $ \infty $, but technically DNE due to sign) |
| 16 | If $ \frac{\tan^2(ax)}{\tan^2(bx)} $, then $ \boxed{\left(\frac{a}{b}\right)^2} $ |
| 17 | If $ \frac{\tan(ax)}{\tan^2(bx)} $, then $ \boxed{\infty} $ (DNE) |
| 18 | If $ \frac{\tan(ax)}{\tan(bx)} $, then $ \boxed{\frac{a}{b}} $ |

But since original says "ab", which is constant, we must conclude:

Corrected Interpretation:

Possibly, "ab" means $ a \cdot b $, but still constant.

So:

- 16: $ \lim_{x \to 0} \frac{\tan^2(ab)}{\tan^2(bx)} = \frac{\text{constant}}{0} \to \infty $ (if $ \tan(ab) \ne 0 $), so $ \infty $ or DNE
- 17: $ \frac{\tan(ab)}{\tan^2(bx)} \to \infty $ (same)
- 18: $ \frac{\tan(ab)}{\tan(bx)} \to \pm\infty $, DNE

But this seems odd.

Alternatively, "ab" is meant to be "ax", typo in formatting.

Given context, most likely typos.

---

## ✔ Final Answers (with correction assumption):

Assume:

- 16: $ \lim_{x \to 0} \frac{\tan^2(ax)}{\tan^2(bx)} = \left(\frac{a}{b}\right)^2 $
- 17: $ \lim_{x \to 0} \frac{\tan(ax)}{\tan^2(bx)} $ → DNE (infinite)
- 18: $ \lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} = \frac{a}{b} $

But since written as "ab", maybe "a" and "b" are constants, and "ab" is product.

But then $ \tan(ab) $ is constant.

So:

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

Let $ c = \tan^2(ab) $, constant.

As $ x \to 0 $, $ \tan(bx) \to 0 $, so $ \tan^2(bx) \to 0^+ $

So:

$$
\frac{c}{\tan^2(bx)} \to \begin{cases}
+\infty & \text{if } c > 0 \\
0 & \text{if } c = 0 \\
-\infty & \text{impossible since } \tan^2 \geq 0
\end{cases}
$$

But $ \tan^2(ab) \geq 0 $, so:

If $ \tan(ab) \ne 0 $, then $ \to +\infty $

If $ \tan(ab) = 0 $, then $ \to 0 $

But without knowing $ a,b $, cannot determine.

So answer depends on $ a,b $

But likely, it's a typo, and should be $ ax $.

---

## ✔ Best Guess (corrected version):

| No | Problem | Answer |
|----|--------|--------|
| 1 | $ \lim_{x\to0} \frac{\sin x}{3x} $ | $ \frac{1}{3} $ |
| 2 | $ \lim_{x\to0} \frac{\sin x}{(1/3)x} $ | $ 3 $ |
| 3 | $ \lim_{x\to0} \frac{\sin 3x}{x} $ | $ 3 $ |
| 4 | $ \lim_{x\to0} \frac{\sin 6x}{2x} $ | $ 3 $ |
| 5 | $ \lim_{x\to0} \frac{\sin x}{3x} $ | $ \frac{1}{3} $ |
| 6 | $ \lim_{x\to0} \frac{3x}{\sin 3x} $ | $ 1 $ |
| 7 | $ \lim_{x\to0} \frac{\tan 2x}{4x} $ | $ \frac{1}{2} $ |
| 8 | $ \lim_{x\to0} \frac{\tan 2x}{\sin 4x} $ | $ \frac{1}{2} $ |
| 9 | $ \lim_{x\to0} \frac{6x}{\tan(x/2)} $ | $ 12 $ |
| 10 | $ \lim_{x\to0} \frac{\sin 3x}{\tan 2x} $ | $ \frac{3}{2} $ |
| 11 | $ \lim_{x\to0} \frac{\sin ax}{\sin bx} $ | $ \frac{a}{b} $ |
| 12 | $ \lim_{x\to0} \frac{\tan ax}{\tan bx} $ | $ \frac{a}{b} $ |
| 13 | $ \lim_{x\to0} \frac{\sin^2 6x}{2x} $ | $ 0 $ |
| 14 | $ \lim_{x\to0} \frac{\sin^2 x}{3x} $ | $ 0 $ |
| 15 | $ \lim_{x\to0} \frac{3x}{\sin^2 3x} $ | DNE (or $ \infty $) |
| 16 | $ \lim_{x\to0} \frac{\tan^2(ax)}{\tan^2(bx)} $ | $ \left(\frac{a}{b}\right)^2 $ |
| 17 | $ \lim_{x\to0} \frac{\tan(ax)}{\tan^2(bx)} $ | DNE |
| 18 | $ \lim_{x\to0} \frac{\tan(ax)}{\tan(bx)} $ | $ \frac{a}{b} $ |

---

## ✔ Final Boxed Answers (assuming corrections):

$$
\boxed{
\begin{aligned}
&1.\ \frac{1}{3} \quad 2.\ 3 \quad 3.\ 3 \quad 4.\ 3 \quad 5.\ \frac{1}{3} \\
&6.\ 1 \quad 7.\ \frac{1}{2} \quad 8.\ \frac{1}{2} \quad 9.\ 12 \quad 10.\ \frac{3}{2} \\
&11.\ \frac{a}{b} \quad 12.\ \frac{a}{b} \quad 13.\ 0 \quad 14.\ 0 \quad 15.\ \text{DNE} \\
&16.\ \left(\frac{a}{b}\right)^2 \quad 17.\ \text{DNE} \quad 18.\ \frac{a}{b}
\end{aligned}
}
$$

> Note: Problems 15–18 likely contain typos. The answers assume that "ab" was meant to be "ax" or similar. If not, then 16–18 involve constants and limits go to infinity or depend on values.