Solve $\lim_{x\to 1}\frac{\sin(x^2-1)}{x-1}$ without L’Hopital.

Question

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

I have tried using squeeze theorem but that doesn’t work. I know you can use derivatives to find that it equals two but is there another to do it?

Answer 1

Hint: Observe, $$\frac{\sin(x^2-1)}{x-1}=\frac{\sin(x^2-1)}{x^2-1}\cdot (x+1)=\frac{\sin y}{y}\cdot (x+1)$$ where $y=x^2-1$. Note, if $x$ tends to $1$ then, $y$ tends to $0$.

Answer 2

$$\lim_{x\to1} \frac{\sin(x^2-1)}{(x-1)}=\lim_{x\to1}\frac{\sin(x^2-1)}{(x^2-1)}\cdot (x+1)=\lim_{x^2\to1}\frac{\sin (x^2-1)}{(x^2-1)}\lim_{x\to1} (x+1)=1\times2$$

Answer 3

Alternatively, you can observe that $\sin(x^{2} - 1) \sim x^{2} - 1$ when $x\to 1$.

Based on such relation, we conclude that \begin{align*} \lim_{x\to 1}\frac{\sin(x^{2} - 1)}{x - 1} & = \lim_{x\to 1}\frac{x^{2} - 1}{x-1}\\\\ & = \lim_{x\to 1}(x+1)\\\\ & = 2 \end{align*}

Answer 4

The above solutions are very simple to understand but this is also what you can do:

Expansion

$\sin(t) = $ $t- \frac{t^3}{3!} +\frac{t^5}{5!}..... $

Now, $\frac{x^2-1}{x-1}-\frac{(x^2-1){3}}{3!(x-1)} +.......$ As ${x\to1} $ all higher powers ${\to0}$

Left with $\frac{(x+1)(x-1)}{x-1}$ As ${x\to1}$ implied $x-1{\to0}$ but is not equal to zero

$$\lim_{x\to1}(x+1){\to2}$$

That is $2$