Question

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

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

Answer 1

Answer: (B), (D)

The correct options are
B is continuous
D is differentiable

we have,

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{x^n}{n!}{\left(\log a\right)}^n=\sum _{n=0}^{\infty }\frac{{\left(x\log a\right)}^n}{n!}$

$=e^{x\log a}=e^{\log a^x}=a^x$

$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{a^{-h}-1}{-h}$

$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{a^h-1}{h}={\log }_{e}a$

Since $L{f}^{\prime }\left(0\right)=R{f}^{\prime }\left(0\right),$

$\therefore f\left(x\right)$ is differentiable at $x=0.$

Since every differentiable function is continuous, therefore, $f\left(x\right)$ is continuous at $x=0.$