The Limit — Question 5

Question 5 - Solution

We want to evaluate: $$\lim_{x \to 0} \frac{\sin(3x)}{\tan(5x)}$$

First, try direct substitution: $$\frac{\sin(3 \cdot 0)}{\tan(5 \cdot 0)} = \frac{\sin(0)}{\tan(0)} = \frac{0}{0}$$

This is an indeterminate form, so we need to rewrite the expression before evaluating the limit.

We will use the standard trigonometric limits: $$\lim_{u \to 0} \frac{\sin u}{u} = 1$$ and $$\lim_{u \to 0} \frac{\tan u}{u} = 1$$

Since $3x \to 0$ and $5x \to 0$ as $x \to 0$, we can apply these limits to $\sin(3x)$ and $\tan(5x)$.

Start with the original expression: $$\frac{\sin(3x)}{\tan(5x)}$$

We want to create the forms $$\frac{\sin(3x)}{3x} \qquad \text{and} \qquad \frac{\tan(5x)}{5x}$$

To do this, multiply and divide by the needed factors: $$\frac{\sin(3x)}{\tan(5x)} = \frac{\sin(3x)}{3x} \cdot \frac{5x}{\tan(5x)} \cdot \frac{3x}{5x}$$

This rewriting is valid because the factors we introduced balance each other out.

Now take the limit of each factor: $$\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1$$

Also, $$\lim_{x \to 0} \frac{5x}{\tan(5x)} = 1$$

This is because $$\lim_{x \to 0} \frac{\tan(5x)}{5x} = 1$$ so its reciprocal also approaches $1$.

Finally, $$\frac{3x}{5x} = \frac{3}{5}$$ for $x \neq 0$. Since limits only depend on values near $0$, not necessarily at $0$, this cancellation is allowed.

Therefore, $$\lim_{x \to 0} \frac{\sin(3x)}{\tan(5x)} = 1 \cdot 1 \cdot \frac{3}{5}$$

$$= \boxed{\frac{3}{5}}$$