Let $\quad f(x)= \begin{cases}(1+a x)^{1 / x} & , x<0 \\ 1+b, & x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2}, & x>0\end{cases}$ be continuous at $x=0$. Then $e^a b c$ is equal to:

<p>$f(x)= \begin{cases}(1+a x)^{1 / x} & , x<0 \\ 1+b & , x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} & , x>0\end{cases}$</p> <p>$$\begin{aligned} &\text { Hence, } f(0)=1+6\\ &\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} \quad\left[\text { For } \frac{0}{0} \text { form, } c=8\right] \end{aligned}$$</p> <p>$$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{2\left(1+\frac{x}{4}\right)^{1 / 2}-2}{\left(1+\frac{x}{8}\right)^{1 / 3}-2} \\ & =\lim _{x \rightarrow 0} \frac{1+\frac{x}{8}-1}{1+\frac{x}{8} \cdot \frac{1}{3}-1}=\frac{\frac{1}{8}}{\frac{1}{8} \cdot \frac{1}{3}}=3 \\ & \therefore 1+b=3 \\ & \Rightarrow b=2 \end{aligned}$$</p> <p>$$\begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0}(1+a x)^{1 / x}=\lim _{x \rightarrow 0} e^{\frac{1+a x-1}{x}}=e^a=3 \\ & \therefore \quad e^a \cdot b c=3 \cdot 2.8=48 \end{aligned}$$</p>