# [SOLVED] Limits

• Mar 26th 2009, 06:43 PM
nzmathman
[SOLVED] Limits
Have come across a limits questions I am having trouble with.

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

$\displaystyle = \lim_{x\to \pi}~\left(\frac{\pi}{\sin{x}} - \frac{x}{\sin{x}}\right)$

I get stuck after that...I know that $\displaystyle \lim_{x\to 0}~\frac{\sin{x}}{x} = 1 ~$but not sure what to do when the limit is for $\displaystyle \pi$, and also the first part becomes undefined which is not the answer - the limit exists. Any tips?

EDIT: sorry I just realised that $\displaystyle \sin{x} = \sin{(\pi-x)}$ and now it is simple - so Q solved!
• Mar 26th 2009, 06:49 PM
mollymcf2009
Quote:

Originally Posted by nzmathman
Have come across a limits questions I am having trouble with.

(1) $\displaystyle \lim_{x\to \pi}~$

$\displaystyle = \lim_{x\to \pi}~\left(\frac{\pi}{\sin{x}} - \frac{x}{\sin{x}}\right)$

I get stuck after that...I know that $\displaystyle \lim_{x\to 0}~\frac{\sin{x}}{x} = 1 ~$but not sure what to do when the limit is for $\displaystyle \pi$, and also the first part becomes undefined which is not the answer - the limit exists. Any tips?

As $\displaystyle x\rightarrow \pi, ~ \frac{\pi-x}{\sin{x}} \rightarrow 0$

Just plug $\displaystyle \pi$ in where you have x and you will see that you have $\displaystyle \frac{0}{0}$
• Mar 26th 2009, 07:08 PM
nzmathman
Actually $\displaystyle \frac{0}{0}$ is the indeterminate form so the expression needs to be "fiddled with" as my lecturer would say....and since I just remembered that $\displaystyle \sin{x} = \sin{(\pi-x)}$, the limit becomes:

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

And then I can apply the rule $\displaystyle \lim_{x\to 0}~\frac{\sin{x}}{x} = 1$
to get the answer of 1.
• Mar 26th 2009, 07:15 PM
mollymcf2009
Quote:

Originally Posted by nzmathman
Actually $\displaystyle \frac{0}{0}$ is the indeterminate form so the expression needs to be "fiddled with" as my lecturer would say....and since I just remembered that $\displaystyle \sin{x} = \sin{(\pi-x)}$, the limit becomes:

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

And then I can apply the rule $\displaystyle \lim_{x\to 0}~\frac{\sin{x}}{x} = 1$
to get the answer of 1.

I always forget about 0/0 as an indeterminate...clear indicator that I need to go to bed!(Sleepy) Thanks for posting your answer!
• Mar 26th 2009, 07:25 PM
nzmathman
Hehe no worries....it's typical that I can't figure out a question so come onto here and finally go and sort out all the latex code for my question only to suddenly see the answer staring me in the face...did the same thing last night but figured it out before I hit submit thread! (Itwasntme) (Thinking)