<s>[INST] Limit: $\lim_{x\to2} \frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)$ Can someone help me understand how the solution for the following limit is $1/4$? 
I've been trying to solve it but I always end up in a 'dead end' with an 
indetermination. If someone could help me, that would be awesome.

$$\lim_{x\to2} \frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)$$

 [/INST] Note that for $x\to 2$
$$\frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)=\frac{\sin\left(\frac{x-2}{x+2}\right)}{x-2}=\frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-2}{x+2}}\cdot\frac{1}{x+2}\to 1\cdot\frac14=\frac14$$
indeed 
$$y=\frac{x-2}{x+2}\to 0 \quad \implies \frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-2}{x+2}}=\frac{\sin y}{y}\to1$$
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