Calculus - Finding limit (NOT L'Hopital's Rule): $\lim_{x \to 1^-}\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}$

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?

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

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 $3$.

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

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 $$f'(1^-)=\left.2x+1+\frac \pi 2 \cos \frac{\pi x}{2} \right|_{x \to 1^-}=3.$$

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

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 $$\lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}=-\frac{\pi}{4}\cdot \lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{\frac{\pi}{4}(1-x)}=-\frac{\pi}{4}.$$

• @Did Thanks a lot! – Robert Z Sep 13 '16 at 5:34

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.