## University Calculus: Early Transcendentals (3rd Edition)

a) $f$ is continuous at $0$ for all $m$. b) For $m=2$, $f$ is differentiable at $0$.
a) For $f$ to be continuous at $x=0$, it is necessary that $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$. We then examine each limit separately. - For $x\to0^-$, since $x\lt0$, we have $f(x)=\sin2x$. Therefore, $\lim_{x\to0^-}f(x)=\sin0=0$ - For $x\to0^+$, since $x\gt0$, we have $f(x)=mx$. Therefore, $\lim_{x\to0^+}f(x)=m\times0=0$ Since $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)$, we see that $f(x)$ is already continuous at $x=0$ no matter which value $m$ might take. b) For $f$ to be differentiable at $x=0$, it is necessary that $\lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)$. We then examine each limit separately. - For $x\to0^-$, since $x\lt0$, we have $f(x)=\sin2x$ Hence, $f'(x)=\cos2x\times(2x)'=2\cos2x$. Therefore, $\lim_{x\to0^-}f'(x)=2\cos(2\times0)=2\cos0=2\times1=2$ - For $x\to0^+$, since $x\gt0$, we have $f(x)=mx$. Hence $f'(x)=m$ Therefore, $\lim_{x\to0^+}f'(x)=m$ It follows that for $\lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)$, $m$ needs to equal $2$. In other words, for $m=2$, $f$ is differentiable at $0$.