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

Question

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?

Answer 1

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 $3$.

Answer 2

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

Answer 3

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

Answer 4

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.