Show this limit is equal to one without using the L'Hôpital rule

Question

Show that

$$\lim_{x\to\pi} \frac{\pi -x}{\sin(x)}=1$$

Although this can be easily solved with L'Hôpital but I need to do it using some algebric simplification. So far I have tried turning $\sin(x)$ to $-\sin(x-\pi)$, breaking into two separate fractions and some of my mates have tried other approaches but no one can figure this out. Can someone point me in the right direction?

$$\pi-x=y$$ https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 — lab bhattacharjee, Dec 12 '19 at 09:44

score 1 · Accepted Answer · answered Dec 12 '19 at 09:51

1

\begin{align}\lim_{x\to\pi}\frac{\pi-x}{\sin x}&=-\frac1{\lim_{x\to\pi}\frac{\sin(x)-\sin(\pi)}{x-\pi}}\\&=-\frac1{\sin'(\pi)}\\&=1\end{align}

answered Dec 12 '19 at 09:51

José Carlos Santos

427,504

1

Nice hat $ \ $ – Maximilian Janisch Dec 12 '19 at 09:52
How did u get sin'(pi)? – Muhammad Usman Dec 12 '19 at 10:06
By definition,$$\sin'(a)=\lim_{x\to a}\frac{\sin(x)-\sin(a)}{x-a}.$$ – José Carlos Santos Dec 12 '19 at 10:10
Sir thanks for the effort but i could accept only one answer and the other one was easier to understand – Muhammad Usman Dec 12 '19 at 10:12

score 0 · Answer 2 · answered Dec 12 '19 at 10:05

0

Since $ \sin(x- \pi)= -\sin x$ we have

$$ \lim_{x \to \pi} \frac{\sin x}{\pi-x}=\lim_{x \to \pi} \frac{\sin ( x - \pi)}{x-\pi}= \lim_{t \to 0} \frac{\sin t}{t}=1.$$

answered Dec 12 '19 at 10:05

Fred

77,394

score 0 · Answer 3 · answered Dec 12 '19 at 14:29

Using the known limit $$\lim_{\color{blue}\theta\to\color{blue}0}\frac{\sin\color{blue}\theta}{\color{blue}\theta},$$ we have

$$\lim_{x\to\pi}\frac{\pi-x}{\sin x}=\lim_{\color{blue}{\pi-x}\to\color{blue}0}\frac{\color{blue}{\pi-x}}{\sin(\color{blue}{\pi-x})}=1.$$

Show this limit is equal to one without using the L'Hôpital rule

3 Answers3