Question

Let \( f(x)=\left\{\begin{array}{cc}e^{-1 / x^{2}} \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right. \) Test continuity and differentiability of \( f(x) \)

Answer 1

\(f(x)= e^{-1/x^2} \sin \frac 1x\)

\(f(0^+) = R.H.D. = L.H.D. = f(0^-)\)

\(f(0^+) = \lim \limits_{h\to 0} \frac{f(h) - f(0)}{h}\)

⇒ \(\lim\limits_{h\to 0} \frac{e^{-1/h^2} \sin \frac 1h - 0}{h}\)

⇒ \(\lim\limits_{h\to 0} \frac{e^{-1/h^2} \sin \frac 1h }{\frac 1h.h.h}\)

⇒ 0

\(f(0^-) = \lim \limits_{h\to 0} \frac{f(-h) - f(0)}{-h}\)

⇒ \(\lim\limits_{h\to 0} \frac{e^{-1/h^2} \sin (-\frac 1h) - 0}{-h. - \frac 1h.(-h)}\)

⇒ \(\lim\limits_{h\to 0} \frac{e^{-1/h^2}}{h^2} \frac{\sin(-\frac 1h)}{(- \frac 1h)}\)

⇒ 0

f(x) is diffrentiable at x.