# Proving $\lim\limits_{x\to a}\frac{1}{x^2-2x}\sin\left(\frac{1}{x}-\frac{1}{x-2}\right)$ does not exist at $a=0,2$

How would I go about proving that given $a=0$ or $a=2$ the following limit does not exist, $$\lim_{x\to a}\frac{\sin\left(\frac{1}{x}-\frac{1}{x-2}\right)}{x^2-2x}$$ Conceptually, the sine component oscillates wildly around that interval and the fractional component $x^2-2x$ seems to approach infinity around $a=0$ and $a=2$. However, I am having difficulty formulating a rigorous proof that the limit does not exist.

• Your argument is correct. The numerator oscillates and denominator goes to $0$. If the limit of fraction existed then using product rule of limits the numerator would also tend to $0$ and not oscillate. Oct 22 '17 at 5:37

Replace $x$ with $\frac{1}{(t-a)}$
So that if $a=0$, replace with $t=\frac{1}{x}$ and if $a=2$, replace with $t = \frac{1}{(x-2)}$ so that our limit changes from ${x\to a}$ to ${t\to \infty}$
Hence our equation becomes (for $a=0$)
$\lim_{t\to \infty}\frac{t\sin\left(t-\frac{t}{1-2t}\right)}{\frac{1}{t}-2}$
Here, value of sin is between $-1$ and $1$ and it is multiplied with a number tending to $\infty$, so it does not exists. Similarly for $a=2$.