Don't Start Hating The L'Hopital's Rule While Solving This Problem! (Part 2)

Rationalizing top and bottom:

$=\large \displaystyle \lim_{x \to 0} \frac{\left(1-\cos ^2x\cos nx\right)}{x^2}\cdot \frac{1}{1+\cos x\sqrt{\cos nx}}$

$=\large \displaystyle \lim_{x \to 0} \frac{\left(1-\cos ^2x\cos nx\right)}{x^2}\cdot \lim_{x \to 0} \frac{1}{1+\cos x\sqrt{\cos nx}}$

$=\large \displaystyle \lim_{x \to 0} \frac{\left(1-\cos ^2x+\cos ^2x-\cos ^2x\cos nx\right)}{x^2}\cdot \frac{1}{1+\cos 0\sqrt{\cos \left(n\cdot 0\right)}}$

$=\large \displaystyle \lim_{x \to 0} \frac{\left(1-\cos ^2x\right)+\cos ^2x\left(1-\cos nx\right)}{x^2}\cdot \frac{1}{1+1\sqrt{\cos \left(0\right)}}$

$=\large \displaystyle \lim_{x \to 0} \left(\frac{1-\cos ^2x}{x^2}+\cos ^2x\frac{\left(1-\cos nx\right)}{x^2}\right)\cdot \frac{1}{1+1\sqrt{1}}$

$=\large \displaystyle \lim_{x \to 0} \left(\frac{\sin ^2x}{x^2}+\cos ^2x\frac{\left(1-\cos nx\right)}{x^2}\cdot \frac{\left(1+\cos nx\right)}{\left(1+\cos nx\right)}\right)\cdot \frac{1}{1+1}$

$=\large \displaystyle \frac{1}{2}\cdot \left(\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2+\lim_{x \to 0} \cos ^2x \cdot \lim_{x \to 0} \frac{1-\cos ^2nx}{x^2}\cdot \frac{1}{\left(1+\cos nx\right)}\right)$

$=\large \displaystyle \frac{1}{2}\cdot \left(1^2+ \cos ^20 \cdot \lim_{x \to 0} \frac{\sin^2 nx}{x^2}\cdot \frac{1}{\left(1+1\right)}\right)$

$=\large \displaystyle \frac{1}{2}\cdot \left(1+ 1 \cdot n^2 \cdot \frac{1}{\left(2\right)}\right)$

$=\large \displaystyle \boxed{\frac{1}{2} (1+\frac{n^2}{2})}$

why not use maclaurin series of cosx about x=0 ? done in 1 step!

JEE Style

Let n = 0 the limit becomes a popular one 1-cosx/x^2 = 1/2

Only one option satisfies

for complete solution we will use L Hopital's rule

Differentiate up and down wrt x

note- i have already put cosnx = 1 and cos = 1 after differentiating you will get

1/2(sinx/x+nsin(nx)/2x)

We know limit x approaching zero sinx/x = 1 and for second limit multiply and divide by n and sin(nx)/nx= 1

we get 1/2(1+n^2/2)