Limit of Trigonometric Functions worksheet - Free Printable

This worksheet contains 18 limit problems, all evaluating limits as x → 0, involving trigonometric functions: sine and tangent. These are classic "special limits" that rely on the fundamental results:

> Key Standard Limits:
>
> 1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
> 2. $\lim_{x \to 0} \frac{\tan x}{x} = 1$

These can be generalized for any constant $k \neq 0$:

> - $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$
> - $\lim_{x \to 0} \frac{\tan(kx)}{kx} = 1$

We will solve each problem using algebraic manipulation to match the standard forms above.

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Solutions:

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1. $\lim_{x\to0} \frac{\sin x}{3x}$

= $\frac{1}{3} \cdot \lim_{x\to0} \frac{\sin x}{x} = \frac{1}{3} \cdot 1 = \boxed{\frac{1}{3}}$

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2. $\lim_{x\to0} \frac{\sin x}{\frac{1}{3}x}$

= $\lim_{x\to0} \frac{\sin x}{x} \cdot 3 = 1 \cdot 3 = \boxed{3}$

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3. $\lim_{x\to0} \frac{\sin 3x}{x}$

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

= $\lim_{u\to0} \frac{\sin u}{u/3} = \lim_{u\to0} 3 \cdot \frac{\sin u}{u} = 3 \cdot 1 = \boxed{3}$

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4. $\lim_{x\to0} \frac{\sin 6x}{2x}$

= $\lim_{x\to0} \frac{\sin 6x}{6x} \cdot \frac{6x}{2x} = 1 \cdot 3 = \boxed{3}$

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5. $\lim_{x\to0} \frac{\sin x}{3x}$ — Same as #1

= $\boxed{\frac{1}{3}}$

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6. $\lim_{x\to0} \frac{3x}{\sin 3x}$

= $\lim_{x\to0} \frac{1}{\frac{\sin 3x}{3x}} = \frac{1}{1} = \boxed{1}$

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7. $\lim_{x\to0} \frac{\tan 2x}{4x}$

= $\lim_{x\to0} \frac{\tan 2x}{2x} \cdot \frac{2x}{4x} = 1 \cdot \frac{1}{2} = \boxed{\frac{1}{2}}$

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8. $\lim_{x\to0} \frac{\tan 2x}{\sin 4x}$

Write as: $\frac{\tan 2x}{2x} \cdot \frac{4x}{\sin 4x} \cdot \frac{2x}{4x}$

= $1 \cdot 1 \cdot \frac{1}{2} = \boxed{\frac{1}{2}}$

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9. $\lim_{x\to0} \frac{6x}{\tan \frac{1}{2}x}$

= $\lim_{x\to0} \frac{6x}{\frac{1}{2}x} \cdot \frac{\frac{1}{2}x}{\tan \frac{1}{2}x} = 12 \cdot 1 = \boxed{12}$

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10. $\lim_{x\to0} \frac{\sin 3x}{\tan 2x}$

= $\lim_{x\to0} \frac{\sin 3x}{3x} \cdot \frac{2x}{\tan 2x} \cdot \frac{3x}{2x} = 1 \cdot 1 \cdot \frac{3}{2} = \boxed{\frac{3}{2}}$

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11. $\lim_{x\to0} \frac{\sin ax}{\sin bx}$

= $\lim_{x\to0} \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b} = 1 \cdot 1 \cdot \frac{a}{b} = \boxed{\frac{a}{b}}$

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12. $\lim_{x\to0} \frac{\tan ax}{\tan bx}$

= $\lim_{x\to0} \frac{\tan ax}{ax} \cdot \frac{bx}{\tan bx} \cdot \frac{a}{b} = 1 \cdot 1 \cdot \frac{a}{b} = \boxed{\frac{a}{b}}$

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13. $\lim_{x\to0} \frac{\sin^2 6x}{2x}$

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

= $\lim_{x\to0} \frac{(\sin 6x)^2}{6x \cdot 6x} \cdot \frac{36x^2}{2x} = \lim_{x\to0} \left( \frac{\sin 6x}{6x} \right)^2 \cdot 18x = 1^2 \cdot 0 = \boxed{0}$

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14. $\lim_{x\to0} \frac{\sin^2 x}{3x}$

= $\lim_{x\to0} \frac{(\sin x)^2}{x^2} \cdot \frac{x^2}{3x} = \lim_{x\to0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{x}{3} = 1 \cdot 0 = \boxed{0}$

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15. $\lim_{x\to0} \frac{3x}{\sin^2 3x}$

= $\lim_{x\to0} \frac{3x}{(3x)^2} \cdot \left( \frac{3x}{\sin 3x} \right)^2 = \lim_{x\to0} \frac{1}{3x} \cdot 1^2 = \infty$

But since we're approaching 0 from both sides, and the expression is positive near 0 (for small x ≠ 0), the limit is positive infinity.

However, in many contexts (especially high school), if the limit diverges, we may say it does not exist or write ∞.

But let’s check:

As $x \to 0$, numerator → 0, denominator → 0, but denominator is squared → goes to zero faster? No!

Actually: $\sin^2 3x \approx (3x)^2 = 9x^2$, so expression ≈ $\frac{3x}{9x^2} = \frac{1}{3x} \to \infty$ as $x \to 0^+$, and $-\infty$ as $x \to 0^-$.

So two-sided limit does not exist.

But since this is likely a basic calculus worksheet expecting finite answers, perhaps there's a typo? Or maybe they expect ∞?

Wait — let me recompute carefully:

$\lim_{x\to0} \frac{3x}{\sin^2 3x} = \lim_{x\to0} \frac{3x}{(3x)^2} \cdot \left( \frac{3x}{\sin 3x} \right)^2 = \lim_{x\to0} \frac{1}{3x} \cdot 1 = \text{undefined (infinite)}$

✔ So answer: Does not exist (or $\infty$ if only right-hand considered). But since x→0, and function is odd? Let’s see:

$f(x) = \frac{3x}{\sin^2 3x}$ → numerator odd, denominator even → overall odd function.

So left limit = -∞, right limit = +∞ → limit does not exist.

But in many Indonesian worksheets, they might accept ∞ if context implies right-hand. However, strictly speaking:

> $\boxed{\text{does not exist}}$

But let’s check problem 16–18 — they have similar structures. Maybe we should assume finite answers? Let’s proceed and come back.

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16. $\lim_{x\to0} \frac{\tan^2 ab}{\tan^2 bx}$ — Wait! This looks suspicious.

If $a$ and $b$ are constants, then $\tan(ab)$ is a constant (unless ab is variable, which it isn't). But $x$ is the variable → $\tan(ab)$ is constant, $\tan(bx)$ → 0 as x→0.

So unless $ab = 0$, this is problematic.

Probably a typo. It should be:

> $\lim_{x\to0} \frac{\tan^2(ax)}{\tan^2(bx)}$

Assuming that’s the case (common in such worksheets):

Then:

= $\left( \lim_{x\to0} \frac{\tan ax}{\tan bx} \right)^2 = \left( \frac{a}{b} \right)^2 = \boxed{\frac{a^2}{b^2}}$

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17. $\lim_{x\to0} \frac{\tan ab}{\tan^2 bx}$ — Again, same issue.

If $ab$ is constant, numerator is constant, denominator → 0 → limit is infinite (unless tan(ab)=0).

Probably meant:

> $\lim_{x\to0} \frac{\tan(ax)}{\tan^2(bx)}$

Then:

= $\lim_{x\to0} \frac{\tan ax}{ax} \cdot \frac{(bx)^2}{\tan^2 bx} \cdot \frac{ax}{(bx)^2} = 1 \cdot 1 \cdot \frac{a}{b^2} \cdot x \to 0$

So = $\boxed{0}$

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18. $\lim_{x\to0} \frac{\tan ab}{\tan bx}$ — again, probably typo.

Should be:

> $\lim_{x\to0} \frac{\tan(ax)}{\tan(bx)} = \boxed{\frac{a}{b}}$ (same as #12)

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## ✔ Final Answers (assuming standard interpretations and fixing likely typos):

| 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 | Does not exist (or ∞ if one-sided) — *but likely expected: $\infty$* |
| 16 | $\frac{a^2}{b^2}$ (assuming typo: tan²(ax)/tan²(bx)) |
| 17 | $0$ (assuming typo: tan(ax)/tan²(bx)) |
| 18 | $\frac{a}{b}$ (assuming typo: tan(ax)/tan(bx)) |

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## 📝 Notes:

- Problems 15, 16, 17, 18 likely contain typographical errors in the original worksheet. The expressions as written don’t make sense because `ab` is treated as a constant while `x` is the variable.
- In educational contexts, especially at high school level, these are almost certainly meant to involve `ax` and `bx`.
- For problem 15, if you must give a numerical answer, and assuming the worksheet expects “∞”, you may write $\infty$. But strictly, the two-sided limit does not exist.

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✔ All problems solved with explanations based on standard trigonometric limit identities.