Question

lim x->0 x^2 sin (1/x)

Answer 1

First of all, we have to find  $\displaystyle\lim_{x\to 0}\ \sin\left(\frac{1}{x}\right).$

Given that  x  tends to 0.  Hence the reciprocal of x, i.e., 1/x, will tend to infinity. We know it.

$x\to 0\ \ \ \Longrightarrow\ \ \ \dfrac{1}{x}\to \infty$

But, even  sin (1/x)  tends to infinity, all possible values of  sin (1/x)  goes to and fro in between [-1, 1]. We can find out from the graph of  sin (1/x).

(Graphs are given as attachments with the answer.)

Hence, we can conclude that,  sin (1/x)  has no limit, so,

$\displaystyle \lim_{x\to 0}\ x^2\sin\left(\dfrac{1}{x}\right)=\lim_{x\to 0}x^2=0^2=\mathbf{0}$

Hence 0 is the answer.