Q.Discuss the continuity of the following function at the point(s) or on the interval indicated against it: $f(x) = \dfrac{\cos 4x - \cos 9x}{1-\cos x}$, for $x \ne 0$, $f(0) = \dfrac{68}{15}$, at $x=0$, on $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$.

$f(x)=\dfrac{\cos4x-\cos9x}{1-\cos x}$ for $x\ne0$, $f(0)=\dfrac{68}{15}$.

Using $\cos A-\cos B=2\sin\left(\dfrac{A+B}2\right)\sin\left(\dfrac{B-A}2\right)$ with $A=4x,B=9x$: $\cos4x-\cos9x=2\sin\left(\dfrac{13x}2\right)\sin\left(\dfrac{5x}2\right)$. Also $1-\cos x=2\sin^2\left(\dfrac x2\right)$.

As $x\to0$: $\sin\left(\tfrac{13x}2\right)\sim\tfrac{13x}2$, $\sin\left(\tfrac{5x}2\right)\sim\tfrac{5x}2$, $\sin^2\left(\tfrac x2\right)\sim\left(\tfrac x2\right)^2$. …

Convert the difference of cosines to a product using $\cos A-\cos B=2\sin\frac{A+B}2\sin\frac{B-A}2$, rewrite $1-\cos x$ …

Assuming the printed constant $68/15$ must be correct and trying to force-fit the algebra to match it, rather than computing the limit independently first and only then comparing …