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

The function is defined as

$$f(x)=\begin{cases}(1+ax)^{1/x}, & x\lt 0\\[4pt]1+b, & x=0\\[4pt]\dfrac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2}, & x\gt 0\end{cases}$$

For continuity at $$x=0$$ we must have

$$\lim_{x\to 0^-}f(x)=f(0)=\lim_{x\to 0^+}f(x).$$

Case 1 : Left-hand limit

Recall the standard limit $$\lim_{x\to 0}\left(1+kx\right)^{\tfrac{1}{x}}=e^{k}.$$ Putting $$k=a$$, we get

$$\lim_{x\to 0^-}(1+ax)^{1/x}=e^{a}.$$

Case 2 : Value at $$x=0$$

From the definition, $$f(0)=1+b.$$ Continuity demands

$$e^{a}=1+b \quad -(1)$$

Case 3 : Right-hand limit

We first make the denominator zero at $$x=0$$ so that a finite limit can exist:

$$\bigl(x+c\bigr)^{1/3}-2=0\ \text{ at }x=0\;\Rightarrow\;c^{1/3}=2\;\Rightarrow\;c=8.$$

Now evaluate the limit with $$c=8$$:

$$L=\lim_{x\to 0^+}\dfrac{(x+4)^{1/2}-2}{(x+8)^{1/3}-2}.$$

Let $$x\to 0^+$$, write $$x=h$$ with $$h\to 0^+$$.

Numerator expansion around $$h=0$$: $$(4+h)^{1/2}=2\left(1+\tfrac{h}{4}\right)^{1/2}\approx 2\left(1+\tfrac{1}{2}\cdot\tfrac{h}{4}\right)=2+\tfrac{h}{4}.$$ Hence $$(4+h)^{1/2}-2\approx\tfrac{h}{4}.$$

Denominator expansion around $$h=0$$: $$(8+h)^{1/3}=2\left(1+\tfrac{h}{8}\right)^{1/3}\approx2\left(1+\tfrac{1}{3}\cdot\tfrac{h}{8}\right)=2+\tfrac{h}{12}.$$ Hence $$(8+h)^{1/3}-2\approx\tfrac{h}{12}.$$

Therefore

$$L=\lim_{h\to 0^+}\dfrac{\tfrac{h}{4}}{\tfrac{h}{12}}=\dfrac{1/4}{1/12}=3.$$

Continuity requires this limit to equal $$f(0)=1+b$$, so

$$1+b=3\;\Rightarrow\;b=2 \quad -(2)$$

Using $$(1)$$ and $$(2)$$:

$$e^{a}=1+b=3\;\Rightarrow\;a=\ln 3.$$

We already obtained $$c=8$$. Finally, compute $$e^{a}\,b\,c$$:

$$e^{a}\,b\,c = 3 \times 2 \times 8 = 48.$$

Thus $$e^{a}bc = 48$$, which corresponds to Option C.