Let f(x)=cases(1+ax)^1/x, & x0cases be continuou…

Question

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

Accepted Answer

RHL: for the $\frac00$ form to resolve nicely, take $c=8$ (so $(x+c)^{1/3}\to2$ as $x\to0$ matching the numerator). Then, writing $(x+4)^{1/2}=2\left(1+\dfrac x4\right)^{1/2}$ and $(x+8)^{1/3}=2\left(1+\dfrac x8\right)^{1/3}$, and using $(1+u)^p-1\approx pu$:

$$\lim_{x\to0^+}\frac{2\left[\left(1+\frac x4\right)^{1/2}-1\right]}{2\left[\left(1+\frac x8\right)^{1/3}-1\right]}=\frac{\frac18}{\frac1{24}}=3$$

So $1+b=3\implies b=2$.

LHL: $\displaystyle\lim_{x\to0^-}(1+ax)^{1/x}=e^a$. By continuity, $e^a=3$ (the common value $f(0)$).

$$e^a\cdot bc=3\cdot2\cdot8=48$$

Solutions · 1