Q.If $f(x) = \dfrac{\sqrt{2+\sin x} - \sqrt3}{\cos^2 x}$, for $x \ne \dfrac{\pi}{2}$, is continuous at $x = \dfrac{\pi}{2}$ then find $f\left(\dfrac{\pi}{2}\right)$.

$f(x)=\dfrac{\sqrt{2+\sin x}-\sqrt3}{\cos^2x}$ for $x\ne\pi/2$, continuous at $\pi/2$, so $f(\pi/2)=\displaystyle\lim_{x\to\pi/2} f(x)$.

Put $x=\pi/2+t$, $t\to0$: $\cos x=-\sin t$, so $\cos^2x=\sin^2t$; and $\sin x=\cos t$, so the numerator is $\sqrt{2+\cos t}-\sqrt3$.

Rationalise the numerator: $\big(\sqrt{2+\cos t}-\sqrt3\big)\big(\sqrt{2+\cos t}+\sqrt3\big)=(2+\cos t)-3=\cos t-1=-2\sin^2(t/2)$.

So $f=\dfrac{-2\sin^2(t/2)}{\sin^2t\left[\sqrt{2+\cos t}+\sqrt3\right]}$. Using $\sin t=2\sin(t/2)\cos(t/2)$, so $\sin^2t=4\sin^2(t/2)\cos^2(t/2)$: …

Substitute $x=\pi/2+t$ to convert $\sin x,\cos x$ to $\cos t,-\sin t$, rationalise the surd numerator, then use the half-angle identity $1-\co …

Trying L'Hospital's rule blindly instead of the algebraic rationalisation — doable, but the half-angle route matches the chapter's tool …