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

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 $2$
