If f(x) = begin{cases} x^2 sin frac{1}{x}, & | vedclass

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(C) To check the continuity of $f(x)$ at $x = 0$,we evaluate the limit as $x 	o 0$.For $x 
eq 0$,$f(x) = x^2 \sin \frac{1}{x}$.We know that $-1 \le

Vedclass · Accepted Answer

$f$ is continuous at $x = 0$

Vedclass · Answer

(C) To check the continuity of $f(x)$ at $x = 0$,we evaluate the limit as $x 	o 0$.For $x 
eq 0$,$f(x) = x^2 \sin \frac{1}{x}$.We know that $-1 \le \sin \frac{1}{x} \le 1$ for all $x 
eq 0$.Multiplying by $x^2$,we get $-x^2 \le x^2 \sin \frac{1}{x} \le x^2$.As $x 	o 0$,both $-x^2 	o 0$ and $x^2 	o 0$.By the Squeeze Theorem,$\lim_{x 	o 0} f(x) = 0$.Since $f(0) = 0$,we have $\lim_{x 	o 0} f(x) = f(0)$.Therefore,$f(x)$ is continuous at $x = 0$.

If $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then

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