Q223 / 270 · NDA 2026 (Phase 1) · Mathematics
What is $\lim\limits_{x\to0}\dfrac{10^{\sin x}-1}{\tan x}$ equal to ?
- A$0$
- B$1$
- Correct answerC$\ln 10$
- D$\log_{10}e$
Solution
The given limit is $\lim_{x\to0}\dfrac{10^{\sin x}-1}{\tan x}$
Multiplying and dividing the expression by $\sin x$:
$\lim_{x\to0}\left(\dfrac{10^{\sin x}-1}{\sin x} \times \dfrac{\sin x}{\tan x}\right)$
$\lim_{x\to0}\left(\dfrac{10^{\sin x}-1}{\sin x} \times \cos x\right)$
Using the standard limit $\lim_{t\to0}\dfrac{a^t-1}{t} = \ln a$, where $t = \sin x \to 0$ as $x \to 0$:
$\ln 10 \times \cos 0$
$\ln 10 \times 1 = \ln 10$
Answer: $\ln 10$
Multiplying and dividing the expression by $\sin x$:
$\lim_{x\to0}\left(\dfrac{10^{\sin x}-1}{\sin x} \times \dfrac{\sin x}{\tan x}\right)$
$\lim_{x\to0}\left(\dfrac{10^{\sin x}-1}{\sin x} \times \cos x\right)$
Using the standard limit $\lim_{t\to0}\dfrac{a^t-1}{t} = \ln a$, where $t = \sin x \to 0$ as $x \to 0$:
$\ln 10 \times \cos 0$
$\ln 10 \times 1 = \ln 10$
Answer: $\ln 10$
Hence, the correct answer is option: 'C'.