Let f(x) = begin{cases} (1+ax)^{1/x} & , x

Question

(C) For $f(x)$ to be continuous at $x=0$,we must have $f(0^-) = f(0) = f(0^+)$.First,$f(0^-) = \lim_{x 	o 0^-} (1+ax)^{1/x} = e^{\lim_{x 	o 0^-} \fr

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) For $f(x)$ to be continuous at $x=0$,we must have $f(0^-) = f(0) = f(0^+)$.First,$f(0^-) = \lim_{x 	o 0^-} (1+ax)^{1/x} = e^{\lim_{x 	o 0^-} \frac{\ln(1+ax)}{x}} = e^a$.Second,$f(0) = 1+b$.Third,for $f(0^+)$ to exist,the denominator must be zero at $x=0$,so $(0+c)^{1/3}-2 = 0 \implies c^{1/3} = 2 \implies c = 8$.Using $L$'$H$ôpital's rule for $f(0^+)$: $\lim_{x 	o 0^+} \frac{\frac{1}{2}(x+4)^{-1/2}}{\frac{1}{3}(x+c)^{-2/3}} = \frac{\frac{1}{2 \sqrt{4}}}{\frac{1}{3} c^{-2/3}} = \frac{1/4}{1/3 \cdot (8)^{-2/3}} = \frac{1/4}{1/3 \cdot 1/4} = 3$.Equating the limits: $e^a = 1+b = 3$.Thus,$e^a = 3$ and $b = 2$.We need to find $e^2bc$. Since $e^a = 3$,$a = \ln 3$. The expression $e^2bc$ is $e^2 \cdot 2 \cdot 8 = 16e^2$. However,re-evaluating the question context,it is likely $e^a bc$. Given the options,$3 \cdot 2 \cdot 8 = 48$.

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 $e^2bc$ is equal to

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