Calculating limit $\lim \frac{1}{x}\sin \frac{1}{x} $

Question

$$\lim \frac{1}{x}\sin \frac{1}{x} = \infty $$ 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.

Answer 1

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.

Answer 2

As a hint , you can take $\frac 1x=a$ so $$\lim_{x \to 0} \frac{1}{x}\sin \frac{1}{x} =\\ \lim_{a \to \infty} \frac{1}{\frac1a}\sin \frac{1}{\frac1a} =\\ \lim_{a \to \infty} a\sin a =?$$