Question

If the function $f\left(x\right)$ defined as $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}3,& x=0\end{array}\\ \begin{array}{cc}{(1+\frac{ax+b{x}^3}{x^2})}^{1/x},& x>0\end{array}\end{array}$ is continuous at $x=0$, then

• A. $a=0$
• B. $b=e^3$
• C. $a=1$
• D. $b=\ln 3$

Answer 1

Answer: (A), (D)

$a=0$
$b=\ln 3$

We have,

$\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{(1+\frac{ah+{bh}^3}{h^2})}^{\frac{1}{h}}=\underset{h\to 0}{lim}{e}^{\frac{1}{h}\ln (1+\frac{ah+{bh}^3}{h^2})}$

For limit to exist, we have

$\underset{h\to 0}{lim}\frac{ah+{bh}^3}{h^2}=0$ i.e., $\underset{h\to 0}{lim}\frac{a+{bh}^2}{h}=0,$

which is possible only if $a=0.$

Now, we have

$\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{(1+bh)}^{\frac{1}{h}}=\underset{h\to 0}{lim}{(1+bh)}^{\frac{b}{bh}}=e^b$

For $f\left(x\right)$ to be continuous at $x=0,$ we have

$\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)\Rightarrow e^b=3$

$\therefore b=\ln 3$