Question

Test the continuity of the function f at x=0, where
$f\left(x\right)=x^2\sin \left(\frac{1}{x}\right)$ for $x\ne 0$
$=1$ for $x=0$

Answer 1

For a function to be continuous at a point, the limiting value of the function at the point must be equal to the value of function at that point.

i.e., $\underset{x\to a}{lim}f\left(x\right)=f\left(a\right)$ ,then f(x) is continuous at x=a.

Here, the function is defined as : $f\left(x\right)=x^2\sin \frac{1}{x}$ ,when $x\ne 0$ and $f\left(x\right)=1$ ,when $x=0$

Thus, to test the continuity of the function at the point x=0, we have to find the limiting value of the function when x$\to$0

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\left\{x^2\sin \frac{1}{x}\right\}$

We know, that value of sin$\theta$ lies in [-1,1] for all values of $\theta$ ,so $\sin \frac{1}{x}$ lies in the interval [-1,1] which implies it is a finite term, let this term be k.

Now, $\underset{x\to 0}{lim}\left\{x^2\sin \frac{1}{x}\right\}=\underset{x\to 0}{lim}\{x^2\times k\}=0^2\times k=0$.

But, according to the defination of the function, $f\left(0\right)=1$.

Thus, $\underset{x\to 0}{lim}\left\{x^2\sin \frac{1}{x}\right\}\ne f\left(0\right)$ which implies that the function is not continuous at x=0.