Let f(x)= cases(1+a x)^1 / x & , x0cases be continuous at x=0.

Question

Let $\quad f(x)= \begin{cases}(1+a x)^{1 / x} & , x0\end{cases}$ be continuous at $x=0$. Then $e^a b c$ is equal to:

Accepted Answer

$f(x)= \begin{cases}(1+a x)^{1 / x} & , x0\end{cases}$

$$\begin{aligned}
&\text { Hence, } f(0)=1+6\
&\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} \quad\left[\text { For } \frac{0}{0} \text { form, } c=8\right]
\end{aligned}$$

$$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{2\left(1+\frac{x}{4}\right)^{1 / 2}-2}{\left(1+\frac{x}{8}\right)^{1 / 3}-2} \
& =\lim _{x \rightarrow 0} \frac{1+\frac{x}{8}-1}{1+\frac{x}{8} \cdot \frac{1}{3}-1}=\frac{\frac{1}{8}}{\frac{1}{8} \cdot \frac{1}{3}}=3 \
& \therefore 1+b=3 \
& \Rightarrow b=2
\end{aligned}$$

$$\begin{aligned}
& \mathrm{LHL}=\lim _{x \rightarrow 0}(1+a x)^{1 / x}=\lim _{x \rightarrow 0} e^{\frac{1+a x-1}{x}}=e^a=3 \
& \therefore \quad e^a \cdot b c=3 \cdot 2.8=48
\end{aligned}$$

Solutions · 1