Find $\lim_{x \rightarrow \pi}\frac{\cos(\frac{\pi}{2}+x)}{x-\pi}$ without using L'Hôpital.

Question

I tried:

$$\lim_{x \rightarrow \pi}\frac{\cos(\frac{\pi}{2}+x)}{x-\pi} = -\frac{\sin(x)}{x-\pi} = ???$$

What do I do next?

Answer 1

$$\lim_{x\to \pi} -\frac{\sin x}{x-\pi} =\lim_{x\to \pi} -\frac{\sin (\pi-x)}{x-\pi} = \lim_{x\to \pi} \frac{\sin (\pi-x)}{\pi-x}$$ Define $t=\pi -x$ then $x\to \pi \Rightarrow t\to 0$ and we get $$\lim_{t\to 0} \frac{\sin t}{t} = 1$$

Answer 2

This is the definition of the derivative at $x=\pi$ of the function $g(x)=\cos(\frac{\pi}2+x)$. (Note that $g(\pi)=0$.) No need for L'Hôpital. If you don't know the chain rule, use a trig identity to write $g(x)=-\sin x$. (Oh, I guess you already did this.) Then it's the derivative of $-\sin(x)$ at $x=\pi$.

(And if you aren't supposed to use calculus at all, then ... convert this to the $\lim\limits_{t\to 0}\dfrac{\sin t}t$ limit.)