Question

Evaluate the limit :
lim x tends to 0+ (ln tan 2x)/(ln tan x) .

Answer 1

Answer on Question # 41152, Math, Integral Calculus

Question:

Evaluate the limit :

$$\lim_{x \text{ tends to } 0+} \frac{\ln \tan 2x}{\ln \tan x}$$

Answer:

$$\lim_{x \to +0} \frac{\ln \tan 2x}{\ln \tan x}$$

Using l'Hôpital's rule:

$$\begin{aligned} \lim_{x \to +0} \frac{(\ln \tan 2x)'}{(\ln \tan x)'} &= \lim_{x \to +0} \frac{\left( \frac{1}{\tan 2x} \cdot \frac{1}{\cos^2 2x} \cdot 2 \right)'}{\left( \frac{1}{\tan x} \cdot \frac{1}{\cos^2 x} \right)'} = 2 \lim_{x \to +0} \frac{\sin x \cos x}{(\sin 2x \cos 2x)} \\ &= 2 \lim_{x \to +0} \frac{\sin 2x}{\sin 4x} \end{aligned}$$

Using $\lim_{x \to +0} \frac{\sin x}{x} = 1$:

$$2 \lim_{x \to +0} \frac{\sin 2x}{\sin 4x} = 2 \frac{2x}{4x} = 1$$

Answer: 1

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