Question

If $f(x)\sum _{n=0}^{\infty }$$\frac{x^n}{n!}{\left(loga\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)\sum _{n=0}^{\infty }$$\frac{x^n}{n!}{\left(loga\right)}^n$$\sum _{n=0}^{\infty }$$\frac{{\left(xloga\right)}^n}{n!}$
$=e^{xloga}=e^{log{a}^x}=a^x$
$L{f}^{\prime }(0)=\underset{h\to 0}{\lim }$ $\frac{f\left(0-h\right)-f\left(0\right)}{-h}=\underset{h\to 0}{\lim }$ $\frac{a^{-h}-1}{-h}=lo{g}_{e}a$
$R{f}^{\prime }(0)=\underset{h\to 0}{\lim }$ $\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }$ $\frac{a^h-1}{h}=lo{g}_{e}a$
Since $=L{f}^{\prime }(0)=R{f}^{\prime }(0),\therefore f(x)$ is differentiable at $x=0$
Since every differentiable function is continuous, therefore, f(x) is continuous at $x=0$