# 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.

• $\sin^2{x}-1\not= \sin{x}(\sin{x}-1)$. Instead, it should be $(\sin{x}-1)(\sin{x}+1)$ – Golbez Aug 23 '14 at 14:04

$$\lim \frac{\sin^2 x-1}{\sin x-1}=\lim_{x \to \frac{\pi}{2}} \frac{(\sin x-1) \cdot (\sin x+1)}{\sin x-1}=\lim_{x \to \frac{\pi}{2}} (\sin x+1)=2$$
Using $a^2-b^2=(a+b)(a-b)$, $$\frac{\sin^2 (x) -1}{\sin (x)-1}=\sin (x)+1$$ The rest is easy.