NIMCET 2027 Limit Previous Year Questions (PYQs) – Page 2 of 3

Given limit is

$\lim_{x\to\infty}\left(\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\right)$

Divide numerator and denominator inside the radical by $x$.

$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$

$=\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

Now,

$\frac{\sqrt{x+\sqrt{x}}}{x}\to 0$ as $x\to\infty$

Therefore,

$\lim_{x\to\infty}\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

$=\sqrt{\frac{1}{1+0}}$

$=1$

We have,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}$

As $x\to 0$,

$|\sin 2x|\sim 2|x|$

Also,

$\log_e(1+|\sin 2x|)\sim |\sin 2x|$

So,

$\log_e(1+|\sin 2x|)\sim 2|x|$

Now the numerator becomes approximately:

$|x|\cdot 2|x|=2x^2$

The denominator becomes approximately:

$x^2(|x|+3)\to 3x^2$

Therefore,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}=\frac{2x^2}{3x^2}$

$=\frac{2}{3}$

Hence, the limit exists and is equal to $\frac{2}{3}$.

Given limit is indeterminate of type $\frac{0}{0}$.

(DL Rule) Differentiate numerator and denominator:

$\displaystyle \lim_{x\to0}\frac{x\tan x}{1-\cos x}=\lim_{x\to0}\frac{\tan x+x\sec^2 x}{\sin x}$

As $x\to0$, $\tan x\approx x$, $\sin x\approx x$, $\sec^2 x\approx1$

$\Rightarrow \dfrac{x+x}{x}=2$

Answer: $\boxed{2}$ ✅