Question. $$y=x^2\sin\left(\frac{1}{x}\right),x\neq0$$ $$y=0,x=0$$ Check Contunity at $x=0$
Answer. In my Text book and videos states that it is continuous. But i got conflicted LHL while solving, i'll show my attemp at the question stepwise. Please look at it and thank you for your reply.
My attempt-
$$lim_{x\rightarrow0}x^2\sin\left(\frac{1}{x}\right)$$ $LHL- Putting\ x=0-h\ \& h\rightarrow0$
$lim_{h\rightarrow0}\left(0-h\right)^2\sin\left(\frac{1}{0-h}\right)$ $=lim_{h\rightarrow0}h^2\sin\left(-\frac{1}{h}\right)$
Now, $\sin\left(-\theta\right)=-\sin\left(\theta\right)$ ,using this
$=lim_{h\rightarrow0}-h^2\sin\left(\frac{1}{h}\right)$
putting value of h whic tends to zero, we will get
$=-0$, where $0$ is approximately close to actual zero
{This is the step where other people answers are different from me, they just ignore the minus signs and write this LHL result =$0$
My RHL is same as of the other people answers.