# limit

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• Apr 26th 2013, 12:41 AM
kastamonu
limit
What is the limit of x->1 sinpi*x/(1-x)^2?
I want to solve it without using L'Hospital rule.
• Apr 26th 2013, 12:48 AM
MarkFL
Re: limit
Hint: use the identity $\displaystyle \sin(\pi-\theta)=\sin(\theta)$ and $\displaystyle \lim_{x\to c}(f(x)\cdot g(x))=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$ and a two-sided limit. What do you find?
• Apr 26th 2013, 01:45 AM
kastamonu
Re: limit
sin(pi-x)*x*(pi-x)/(1-X)(1+x)(pi-x)=x*(pi-x)/(1-x)(1+x)?
• Apr 26th 2013, 01:52 AM
MarkFL
Re: limit
No...I'm sorry but that makes no sense to me...please read my post above again and apply the identities I gave...
• Apr 26th 2013, 04:13 AM
kastamonu
Re: limit
which one is f(x) or g(x)?
• Apr 27th 2013, 12:46 AM
kastamonu
Re: limit
• Apr 27th 2013, 01:51 AM
MarkFL
Re: limit
We are given to evaluate:

$\displaystyle \lim_{x\to1}\frac{\sin(\pi x)}{(1-x)^2}$

The first thing I am looking for is a way to apply the well-known result:

(1) $\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1$

Seeing the denominator has $\displaystyle 1-x$ as a factor, this takes me to $\displaystyle \sin(\pi x)=\sin(\pi-\pi x)=sin(\pi(1-x))$. So we now have:

$\displaystyle \lim_{x\to1}\frac{\sin(\pi(1-x))}{(1-x)^2}$

Now, let's use the substitution $\displaystyle x=1-x$ and we have:

$\displaystyle \lim_{u\to0}\frac{\sin(\pi u)}{u^2}$

Now let's finish rewriting the expression so we can apply (1):

$\displaystyle \lim_{u\to0}\frac{\pi\sin(\pi u)}{\pi u^2}=\lim_{u\to0}\frac{\sin(\pi u)}{\pi u}\cdot\frac{\pi}{u}=\lim_{u\to0}\frac{\sin(\pi u)}{\pi u}\cdot\lim_{u\to0}\frac{\pi}{u}$

Applying (1), we are left with:

$\displaystyle \pi\lim_{u\to0}\frac{1}{u}$

Now, do we have:

$\displaystyle \lim_{u\to0^{-}}\frac{1}{u}=\lim_{u\to0^{+}}\frac{1}{u}$ ?
• Apr 27th 2013, 09:40 AM
kastamonu
Re: limit
sin(pi*x) sin(pi*(x-1)+pi)
--------- = ----------------
(1-x)^2 (1-x)^2
we can replace (1-x) with u.
I am stuck in here.
• Apr 27th 2013, 10:05 AM
MarkFL
Re: limit
You have incorrectly applied the trigonometric identity.
• Apr 27th 2013, 10:10 PM
kastamonu
Re: limit
-sin(pi*(x-1))
= --------------
(1-x)^2

sin(pi*(1-x))
= ------------- ( since -sin(y) = sin(-y) )
(1-x)^2
• Apr 27th 2013, 10:18 PM
MarkFL
Re: limit
Okay, but that's really unnecessary...did you actually read my post above where I showed you what to do except for the trivial last step? I hate to think my efforts are in vain. :D
• Apr 27th 2013, 10:27 PM
kastamonu
Re: limit
I couldn't get it.I am sorry.
• Apr 27th 2013, 10:33 PM
MarkFL
Re: limit
What exactly about what I posted don't you get...I want to help, but my main objective is for you to truly understand...tell me where I lose you in my post...
• Apr 27th 2013, 10:44 PM
MarkFL
Re: limit
edit: web error...sorry.
• Apr 27th 2013, 11:27 PM
kastamonu
Re: limit