Calculating limit $\lim \frac{1}{x}\sin \frac{1}{x} $ $$\lim \frac{1}{x}\sin \frac{1}{x} = \infty $$ when $x \to 0$
But by the graph I doubt we should get an infinite oscillating discontinuity and so the limit does not exist here. But I am unable to point out the flaw. Please help or suggest.

Let
$$f(x)=\frac 1x\sin(\frac 1x)$$
$$u_n=\frac{1}{n\pi}$$
$$v_n=\frac{1}{\frac{\pi}{2}+2n\pi}$$
then
$$\lim_{n\to+\infty}u_n=\lim_{n\to+\infty}v_n=0$$
$$\lim_{n\to+\infty}f(u_n)=0$$
$$\lim_{n\to+\infty}f(v_n)=+\infty$$
can you conclude.
