Let $f(x) = \begin{cases} (1+ax)^{1/x} & ,\ x < 0 \\ 1+b & ,\ x = 0 \\ \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & ,\ x > 0 \end{cases}$ be continuous at $x = 0$. Then $e^a bc$ is equal to
- (1) $64$
- (2) $72$
- (3) $48$
- (4) $36$
Solution:
Since $f(x)$ is continuous at $x=0$, the left-hand limit (LHL), right-hand limit (RHL), and value of the function at $x=0$ must be equal.
$$f(0^-) = \lim_{x \to 0^-} (1+ax)^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{ax}{x}} = e^a$$
$$f(0) = 1 + b$$
$$f(0^+) = \lim_{x \to 0^+} \frac{(x+4)^{1/2} – 2}{(x+c)^{1/3} – 2}$$
For the limit to exist finitely, the denominator must approach $0$ as $x \to 0$ because the numerator approaches $0$.
$$c^{1/3} – 2 = 0 \Rightarrow c^{1/3} = 2 \Rightarrow c = 8$$
Applying L’Hôpital’s Rule to find $f(0^+)$:
$$f(0^+) = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{x+4}}}{\frac{1}{3}(x+8)^{-2/3}}$$
$$f(0^+) = \frac{\frac{1}{2(2)}}{\frac{1}{3}(8)^{-2/3}} = \frac{\frac{1}{4}}{\frac{1}{3 \times 4}} = \frac{3}{4} \times 4 = 3$$
Equating $f(0^-) = f(0) = f(0^+)$:
$$e^a = 1 + b = 3$$
$$e^a = 3$$
$$1 + b = 3 \Rightarrow b = 2$$
$$c = 8$$
Evaluating the required expression:
$$e^a b c = 3 \times 2 \times 8 = 48$$
Ans. (3)