JEE Main 2023 — Limits Question with Solution

Given,

$x=2$ be the root of the given equation $x^2+px+q=0$,

Putting $x=2$ in given equation we get,

$\Rightarrow 4+2p+q=0\Rightarrow q+4=-2p$

$\because x^2-4px+q^2+8q+16$

$=x^2-4px+{\left(q+4\right)}^2$

$=x^2-4px+4p^2 (\because q+4=-2p)$

$=(x-2p{)}^2$

Now, solving the limit $\underset{x\to 2p^{+}}{\lim }\left[f\left(x\right)\right]=\underset{x\to 2p^{+}}{\lim }\left[\frac{1-\cos \left({\left(x-2p\right)}^2\right)}{{\left(x-2p\right)}^4}\right]$

Let $x-2p=\theta$

$\Rightarrow \underset{\theta \to 0^{+}}{\lim }\left[f\left(x\right)\right]=\underset{\theta \to 0^{+}}{\lim }\left[\frac{1-\cos \left({\theta }^2\right)}{{\left(\theta \right)}^4}\right]$

$\because \underset{\theta \to 0}{\lim }\left(\frac{1-\cos \left({\theta }^2\right)}{{\left(\theta \right)}^4}\right)=\frac{1}{2}$   (Using L'Hospital's)

$\therefore \underset{\theta \to 0^{+}}{\lim }\left[f\left(x\right)\right]=\underset{\theta \to 0^{+}}{\lim }\left[\frac{1}{2}\right]$

$\therefore \underset{\theta \to 0^{+}}{\lim }\left[f\left(x\right)\right]=0$