Question

\(\lim_{x \to 0}\) \(\frac{3^{\sin x} - 2^{\tan x}}{\sin x}\) =

• A. 0
• B. 1
• C. \(\log_e 6\)
• D. \(\log_e \frac{3}{2}\)

Hint:

Always recheck initial conditions before applying L'Hôpital's Rule to confirm if the limit can be directly calculated.

Answer 1

The Correct Option is D

Let $L = \lim_{x \to 0} \frac{3^{\sin x} - 2^{\tan x}}{\sin x}$. We can rewrite the expression as: $$L = \lim_{x \to 0} \frac{3^{\sin x} - 1 - (2^{\tan x} - 1)}{\sin x}$$ $$L = \lim_{x \to 0} \frac{3^{\sin x} - 1}{\sin x} - \lim_{x \to 0} \frac{2^{\tan x} - 1}{\sin x}$$ We know that $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$. So, $\lim_{x \to 0} \frac{3^{\sin x} - 1}{\sin x} = \ln 3$. For the second term: $$\lim_{x \to 0} \frac{2^{\tan x} - 1}{\sin x} = \lim_{x \to 0} \frac{2^{\tan x} - 1}{\tan x} \cdot \frac{\tan x}{\sin x}$$ $$\lim_{x \to 0} \frac{2^{\tan x} - 1}{\tan x} = \ln 2$$ $$\lim_{x \to 0} \frac{\tan x}{\sin x} = \lim_{x \to 0} \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} = \lim_{x \to 0} \frac{1}{\cos x} = 1$$ Thus, $\lim_{x \to 0} \frac{2^{\tan x} - 1}{\sin x} = \ln 2 \cdot 1 = \ln 2$. Therefore, $$L = \ln 3 - \ln 2 = \ln \frac{3}{2}$$ Final Answer: The final answer is $\boxed{(4)}$