Finding the limit of a trigonometric function $\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$ Could someone please clarify whether my calculation on the following limit problem is correct?
Determine the following limit:
$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$
$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$ = $\lim_{x \to \frac{\pi}{2}} \sin x\frac{\sin x-1}{\sin x-1} = \lim_{x \to \frac{\pi}{2}}\sin x \frac{1}{1} = \lim_{x \to \frac{\pi}{2}} \sin x = 1$
Thank you.

$$\lim \frac{\sin^2 x-1}{\sin x-1}=\lim_{x \to \frac{\pi}{2}} \frac{(\sin x-1) \cdot (\sin x+1)}{\sin x-1}=\lim_{x \to \frac{\pi}{2}} (\sin x+1)=2$$
