Let x = 2 be a root of the equation x ^ 2 + px + q = 0 and f(x) = \...

Given,

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

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

$⇒ 4 + 2p + q = 0 ⇒ q + 4 = - 2p$

$∵ {x}^{2} - 4px + {q}^{2} + 8q + 16$

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

$= {x}^{2} - 4px + 4{p}^{2} ( ∵ q + 4 = - 2p )$

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

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

Let $x - 2p = \theta$

$⇒ \lim_{\theta \to {0}^{+}}\left[f\left(x\right)\right] = \lim_{\theta \to {0}^{+}}\left[\dfrac{1 - \cos \left({\theta}^{2}\right)}{{\left(\theta\right)}^{4}}\right]$

$∵ \lim_{\theta \to 0}\left(\dfrac{1 - \cos \left({\theta}^{2}\right)}{{\left(\theta\right)}^{4}}\right) = \dfrac{1}{2}$ (Using L'Hospital's)

$∴ \lim_{\theta \to {0}^{+}}\left[f\left(x\right)\right] = \lim_{\theta \to {0}^{+}}\left[\dfrac{1}{2}\right]$

$∴ \lim_{\theta \to {0}^{+}}\left[f\left(x\right)\right] = 0$