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

Question

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.

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

Answer 1

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

$$y=\frac{x-2}{x+2}\to 0 \quad \implies \frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-2}{x+2}}=\frac{\sin y}{y}\to1$$

Answer 2

You can use $\sin(x)\sim x$ when $x$ tends to $0$. It basically means that you can remove the $\sin$ in your expression. Is it easier then?

Another approach, this limit is by definition the derivative of the function $$g(x)=\sin\left(\dfrac{x-2}{x+2}\right)$$ at $x=2$. If you can compute the derivative of $g$ using the chain rule, all you have to do is replace $x$ with $2$ afterwards.