How to answer Calculus by Michael Spivak Chapter 5 problem 15.IX

Question

I have had troubles with the problem 15.

I have to evaluate $$\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1}$$

Using the number:

$$\alpha = \lim_{x\to 0} \frac{\sin(x)}{x}$$

I don't have to use L'Hopital because the chapter is about limits only, I know the answer must be 2, here it is what I did and I want to know what I did wrong:

$\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1} = \lim_{x\to 1} \frac{\sin(x^2-1)(x+1)}{x^2-1}$

Say $y = x^2-1$

$ x= \pm \sqrt{y+1}$

Then

$\lim_{y\to 0} \frac{\sin(y)(\pm\sqrt{y+1}+1)}{y} = \alpha * \lim_{y\to 0} \pm\sqrt{y+1}+1 = ??$

I don't know if I had to do the $\pm$.

I apologize if it a silly mistake.

Answer 1

All you have to do is$$\lim_{x\to1}\frac{\sin(x^2-1)}{x-1}=2\lim_{x\to1}\frac{\sin(x^2-1)}{(x-1)(x+1)}=2\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}=2\alpha=2.$$

Answer 2

Beside the nice and simple solution José Carlos Santos provided, let $x=1+y$ with $y\to 0$. This makes $$\lim_{x\to1}\frac{\sin(x^2-1)}{x-1}=\lim_{y\to 0}\frac{\sin (y (y+2))}{y}$$ Using equivalents $y(y+2)\sim 2y$ making $$\frac{\sin (y (y+2))}{y}\sim \frac{\sin(2y)} y=2\frac{\sin(2y)} {2y}=2\frac{\sin(z)} {z}$$

If you want more, use the Taylor series to get $$\frac{\sin (y (y+2))}{y}=2+y+O\left(y^2\right)$$ which shows the limit and how it is approached.

You could also expand $$\sin (y (y+2))=\sin(y^2)\cos(2y)+\sin(2y) \cos(y^2)$$ and notice that both cosines tend to $1$ and rewrite $$\frac{\sin (y (y+2))}{y}=y \frac{\sin(y^2)}{y^2} \cos(2y)+2\frac{\sin(2y)}{2y}\cos(y^2)$$