Q.Discuss the continuity of $f(x)$ at $x = \dfrac{\pi}{4}$ where, $f(x) = \dfrac{(\sin x + \cos x)^3 - 2\sqrt2}{\sin 2x - 1}$, for $x \ne \dfrac{\pi}{4}$, $= \dfrac{3}{\sqrt2}$, for $x = \dfrac{\pi}{4}$.

$f(x)=\dfrac{(\sin x+\cos x)^3-2\sqrt2}{\sin2x-1}$ for $x\ne\pi/4$, $f(\pi/4)=\dfrac{3}{\sqrt2}$.

Put $x=\pi/4+t$, $t\to0$. Then $\sin x+\cos x=\sqrt2\cos t$ (since $\sin x+\cos x=\sqrt2\sin(x+\pi/4)=\sqrt2\cos t$ after the shift), so $(\sin x+\cos x)^3=2\sqrt2\cos^3t$, and the numerator is $2\sqrt2\cos^3t-2\sqrt2=2\sqrt2(\cos^3t-1)$.

Also $\sin2x=\sin(\pi/2+2t)=\cos2t$, so the denominator is $\cos2t-1=-2\sin^2t$.

Using $1-\cos^3t=(1-\cos t)(1+\cos t+\cos^2t)$, with $1-\cos t\sim t^2/2$ and $1+\cos t+\cos^2t\to3$ as $t\to0$, while $\sin^2t\sim t^2$: …

Substitute $x=\pi/4+t$ so $\sin x+\cos x$ collapses to $\sqrt2\cos t$ and $\sin2x$ becomes $\cos2t$, then use the factorisation $1-\cos^3t=(1- …

Expanding $(\sin x+\cos x)^3$ directly with the binomial theorem in terms of $\sin x,\cos x$ instead of first collapsing it to …