<s>[INST] Limit involving a trigonometric function as $x\rightarrow +\infty$ I am probably supposed to use this identity: $$\lim_{x\to0} \frac{\sin(x)}{x} = 1$$ 
I have this mathematical problem:
$$\lim_{x\to+∞} (x^2-1) \sin(\frac{1}{x-1}) = \lim_{x\to+∞} (x+1)(x-1) \sin(\frac{1}{x-1}) = ???$$
Can I do this to the sinus limit?
$$\lim_{x\to+∞} \frac{1}{\sin(\frac{x-1}{1})}$$
 [/INST] Let $X=\frac{1}{x-1}$ so that $x-1=\frac{1}{X}$ and $x+1=2+\frac{1}{X}$
We have $\lim_{x \to +\infty} (x^2-1) \sin(\frac{1}{x-1})=\lim_{X \to 0^+} (2+\frac{1}{X}) \frac{\sin(X)}{X}=+\infty$
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