Question

question 20 find the limit. \\( \lim_{x \to 0} \frac{\tan(4x)}{x} = \\)

Explanation:

Step1: Rewrite tan(4x)

$\tan(4x) = \frac{\sin(4x)}{\cos(4x)}$, so the limit becomes:
$$\lim_{x \to 0} \frac{\sin(4x)}{x\cos(4x)}$$

Step2: Adjust for standard limit

Multiply numerator and denominator by 4:
$$\lim_{x \to 0} \frac{4\sin(4x)}{4x\cos(4x)} = 4 \cdot \lim_{x \to 0} \frac{\sin(4x)}{4x} \cdot \lim_{x \to 0} \frac{1}{\cos(4x)}$$

Step3: Evaluate standard limits

We know $\lim_{u \to 0} \frac{\sin(u)}{u}=1$ (let $u=4x$, $u\to0$ as $x\to0$) and $\lim_{x \to 0} \cos(4x)=\cos(0)=1$:
$$4 \cdot 1 \cdot \frac{1}{1} = 4$$

Answer:

$4$