Question

If $f (x) \displaystyle \sum_{n=0}^\infty$$\frac{x^{n}}{n!}\left(log\,a\right)^{n},$ then at $x = 0, f(x)$

• A. has no limit
• B. is discontinuous
• C. is continuous but not differentiable
• D. is differentiable

Answer 1

Answer: (D)

We have, $f (x) \displaystyle \sum_{n=0}^\infty$$\frac{x^{n}}{n!}\left(log\,a\right)^{n}$$\displaystyle \sum_{n=0}^\infty$$\frac{\left(x\,log\,a\right)^{n}}{n!}$
$=e^{x\,log\,a}=e^{log\,a^x}=a^{x}$
$Lf '(0)=\displaystyle \lim_{h \to 0}$ $\frac{f \left(0-h\right)-f \left(0\right)}{-h}=\displaystyle \lim_{h \to 0}$ $\frac{a^{-h}-1}{-h}=log_e\,a$
$Rf '(0)=\displaystyle \lim_{h \to 0}$ $\frac{f \left(0+h\right)-f \left(0\right)}{h}=\displaystyle \lim_{h \to 0}$ $\frac{a^{h}-1}{h}=log_e\,a$
Since $= Lf '(0) =Rf' (0), \,\therefore f(x)$ is differentiable at $x = 0$
Since every differentiable function is continuous, therefore, f(x) is continuous at $x = 0$