4
$\begingroup$

How do I find this limit?

$$\displaystyle{\lim_{x \to 1^-}}\frac{x^2+x+\sin({\pi \over 2}x)-3}{x-1}$$

I am unable to factor the numerator to get rid of the denominator. Can someone please help? Thank you!

Is there any other way to get the answer besides using L'Hopital's Rule?

$\endgroup$
  • 1
    $\begingroup$ what kind of rules can you use? $\endgroup$ – Dr. Sonnhard Graubner Sep 12 '16 at 15:04
  • $\begingroup$ Are you sure the factor is correct? It's even not an indeterminate form 0/0, the numerator tends to $-1+\sin 0.5$ $\endgroup$ – Denis Korzhenkov Sep 12 '16 at 15:05
  • $\begingroup$ Did you mean $\sin(\pi x/2)$? $\endgroup$ – Olivier Oloa Sep 12 '16 at 15:06
  • 1
    $\begingroup$ Yes I meant sin(πx/2)! Sorry for the confusion! $\endgroup$ – user1234 Sep 12 '16 at 15:11
  • $\begingroup$ You can find the limit easily using Hôpital's rule. $\endgroup$ – user296113 Sep 12 '16 at 15:12
6
$\begingroup$

Note that we can write

$$\begin{align} \frac{x^2+x+\sin(\pi x/2)-3}{x-1}&=(x+2)+\frac{\sin(\pi x/2)-1}{x-1}\\\\ &=(x+2)-2\frac{\sin^2(\pi(x-1)/4)}{x-1}\\\\ &=(x+2)-\frac{\pi^2}{8}\left(\frac{\sin(\pi(x-1)/4)}{\pi (x-1)/4}\right)^2(x-1) \end{align}$$

Inasmuch as $\lim_{x\to 1^-}\frac{\sin(\pi(x-1)/4)}{\pi (x-1)/4}=1$, the limit of interest is $3$.

$\endgroup$
  • 3
    $\begingroup$ Yeah ! Real mathematicians NEVER use l'hospitals's rule. $\endgroup$ – Rene Schipperus Sep 12 '16 at 15:30
  • $\begingroup$ @ReneSchipperus Rene, now that was funny! And much appreciative. -Mark $\endgroup$ – Mark Viola Sep 12 '16 at 15:31
  • $\begingroup$ @Dr.MV how did you get from the first step to the second step (the part with sin^2)? thanks so much! $\endgroup$ – user1234 Sep 12 '16 at 16:17
  • $\begingroup$ @Rachel Note that $$\sin^2(x/2)=\frac{1-\cos(x)}{2}$$ $\endgroup$ – Mark Viola Sep 12 '16 at 17:19
7
$\begingroup$

One may recall that, for any differentiable function $f$ near $a$, one has $$ \lim_{x \to a^-}\frac{f(x)-f(a)}{x-a}=f'(a^-) $$ then observing that $$ \frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}=\frac{\left(x^2+x+\sin({\pi\over 2}x)\right)-\left(1^2+1+\sin({\pi\over 2}\cdot1)\right)}{x-1} $$ one gets that the sought limit is equal to $$ f'(1^-)=\left.2x+1+\frac \pi 2 \cos \frac{\pi x}{2} \right|_{x \to 1^-}=3. $$

$\endgroup$
  • $\begingroup$ such a funny hint! $\endgroup$ – Denis Korzhenkov Sep 12 '16 at 15:15
  • $\begingroup$ As I always tell my students, when you know derivatives, you know many limits. But probably the OP's instructor won't agree. $\endgroup$ – egreg Sep 12 '16 at 16:38
1
$\begingroup$

Here we do not use Hopital or derivatives.

Note that $$x^2+x+\sin({\pi \over 2}x)-3=(x+2)(x-1)+\sin({\pi \over 2}-{\pi \over 2}(1-x))-1\\=(x+2)(x-1)+\cos({\pi \over 2}(1-x))-1\\ =(x+2)(x-1)-2\sin^2(\frac{\pi}{4}(1-x))$$ because $1-\cos(t)=2\sin^2(t/2)$. Hence, as $x\to 1$, $$\frac{x^2+x+\sin({\pi \over 2}x)}{(x-1)}=x+2-2\sin(\frac{\pi}{4}(1-x))\cdot \frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}\to 1+2+0=3$$ where we used the fact that $$\lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}=-\frac{\pi}{4}\cdot \lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{\frac{\pi}{4}(1-x)}=-\frac{\pi}{4}.$$

$\endgroup$
  • $\begingroup$ @Did Thanks a lot! $\endgroup$ – Robert Z Sep 13 '16 at 5:34
0
$\begingroup$

Another possible way to do it.

Start changing variable $x=y+1$; this gives $$\frac{x^2+x+\sin \left(\frac{\pi x}{2}\right)-3}{x-1}=\frac{y^2+3 y+\cos \left(\frac{\pi y}{2}\right)-1}{y}$$ Now, use Taylor expansion $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ which gives $$\cos \left(\frac{\pi y}{2}\right)=1-\frac{\pi ^2 y^2}{8}+O\left(y^4\right)$$ So $$\frac{y^2+3 y+\cos \left(\frac{\pi y}{2}\right)-1}{y}=\frac{y^2 +3y-\frac{\pi ^2 y^2}{8}+O\left(y^4\right)}{y}=(1-\frac{\pi ^2 }{8})y+3+O\left(y^3\right)$$ which shows the limit and how it is approached.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.