NIMCET 2026 — Mathematics PYQ

To evaluate the limit, we analyze the behavior of the expression as $x$ approaches $0$ from both the left ($x\to 0^{-}$) and the right ($x\to 0^{+}$).

1. Using Standard Limits

Recall that ${\lim }_{u\to 0}\frac{{\log }_e(1+u)}{u}=1$ and ${\lim }_{x\to 0}\frac{\sin 2x}{2x}=1$.

The expression can be rewritten by manipulating the terms:

$$\frac{|x|{\log }_e(1+|\sin 2x|)}{x^2(|x|+3)}=\frac{|x|}{x^2}\cdot \frac{{\log }_e(1+|\sin 2x|)}{|\sin 2x|}\cdot \frac{|\sin 2x|}{(|x|+3)}$$

Since $\frac{|x|}{x^2}=\frac{|x|}{|x{|}^2}=\frac{1}{|x|}$, we have:

$$\underset{x\to 0}{\lim }\left(\frac{1}{|x|}\cdot \frac{{\log }_e(1+|\sin 2x|)}{|\sin 2x|}\cdot \frac{|\sin 2x|}{|x|+3}\right)$$

2. Evaluating the Limit

As $x\to 0$, we know $|\sin 2x|\to 0$, so $\frac{{\log }_e(1+|\sin 2x|)}{|\sin 2x|}\to 1$.

The expression simplifies to:

$$\underset{x\to 0}{\lim }\frac{|\sin 2x|}{|x|(|x|+3)}$$

Since $|\sin 2x|\approx |2x|$ for small $x$:

$$\underset{x\to 0}{\lim }\frac{2|x|}{|x|(|x|+3)}=\underset{x\to 0}{\lim }\frac{2}{|x|+3}$$

3. Final Calculation

Substitute $x=0$ into the simplified expression:

$$\frac{2}{0+3}=\frac{2}{3}$$

Since the limit from the left and the right both converge to $\frac{2}{3}$, the limit exists.

Conclusion: The correct option is (b).