NDA 2026 — Mathematics PYQ

The given limit is:

$$\underset{x\to 0}{\lim }\frac{10^{\sin x}-1}{\tan x}$$

If we substitute $x=0$ directly, we get $\frac{10^0-1}{\tan 0}=\frac{1-1}{0}=\frac{0}{0}$, which is an indeterminate form. We can solve this by rearranging the expression into standard limit forms.

1. Algebraic Manipulation:

Multiply and divide by $\sin x$:

$$\underset{x\to 0}{\lim }\left(\frac{10^{\sin x}-1}{\sin x}\cdot \frac{\sin x}{\tan x}\right)$$

2. Simplifying the components:

• Recall the standard limit: ${\lim }_{u\to 0}\frac{a^u-1}{u}=\ln a$. Here, $u=\sin x$, and as $x\to 0$, $u\to 0$. Thus:

  $$\underset{x\to 0}{\lim }\frac{10^{\sin x}-1}{\sin x}=\ln 10$$

• Simplify the second part:

  $$\frac{\sin x}{\tan x}=\frac{\sin x}{\frac{\sin x}{\cos x}}=\cos x$$

  As $x\to 0$, $\cos x\to 1$.

3. Combining the results:

$$\left(\underset{x\to 0}{\lim }\frac{10^{\sin x}-1}{\sin x}\right)\cdot \left(\underset{x\to 0}{\lim }\cos x\right)$$

$$=(\ln 10)\cdot (1)$$

$$=\ln 10$$

Final Answer:

The correct option is (c) $\ln 10$.