JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Limits

As $x\to 0$, both numerator and denominator approach $8+8+3\sin 2=16+3\sin 2$, giving the indeterminate form $1^{\infty }$.

Using the standard limit ${\lim }_{x\to 0}(1+f(x){)}^{g(x)}=e^{{\lim }_{x\to 0}f(x)\cdot g(x)}$:

Let $f(x)=\frac{N(x)-D(x)}{D(x)}$ where $N$ is the numerator and $D$ is the denominator.

We need ${\lim }_{x\to 0}\frac{100}{x}\cdot \frac{N(x)-D(x)}{D(x)}$.

Expanding $N(x)-D(x)$ using Taylor series around $x=0$:

• $(x+2\cos x)\approx 2-\frac{x^2}{2}$ and $(x+2)=2+x$
• The difference in the cubic and quadratic terms, plus the sine terms, when expanded to first order in $x$, gives terms that cancel.

Applying L'Hôpital's rule carefully to the ratio $\frac{N(x)-D(x)}{x}$ as $x\to 0$ yields 0.

Therefore, the limit is $e^0=1$.