Q.Discuss the continuity of the following function at the point(s) or on the interval indicated against it: $f(x) = \dfrac{\sin^2 \pi x}{3(1-x)^2}$, for $x \ne 1$, $= \dfrac{\pi^2 \sin^2\left(\frac{\pi x}{2}\right)}{3+4\cos^2\left(\frac{\pi x}{2}\right)}$ for $x=1$, at $x=1$.

$f(x)=\dfrac{\sin^2(\pi x)}{3(1-x)^2}$ for $x\ne1$, and $f(1)=\dfrac{\pi^2\sin^2(\pi x/2)}{3+4\cos^2(\pi x/2)}$ evaluated at $x=1$: since $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$, this gives $f(1)=\dfrac{\pi^2(1)}{3+0}=\dfrac{\pi^2}{3}$.

For the limit, put $x=1+h$, $h\to0$: $\sin(\pi x)=\sin(\pi+\pi h)=-\sin(\pi h)$, so $\sin^2(\pi x)=\sin^2(\pi h)$; and $(1-x)^2=h^2$. …

Evaluate $f(1)$ directly from the given alternate formula (using $\sin(\pi/2)=1,\cos(\pi/2)=0$), then compute the ordinary limit of the $x\ne1$ branch via the …

Not realising the value 'for $x=1$' is itself given as a formula to be evaluated (not a bare number) — it must first be simplified using $\sin(\pi/2)=1$ a …