Given the function $$ f(x) = \begin{cases} (1 + 2^{\frac{3}{x}})^{bsin(x)} &\quad if \quad x\gt 0 \\ \\ \frac{arctan(9bx)}{x} &\quad if \quad x\lt 0 \\ \end{cases} $$
Prove that exist $b \gt 0$, so that $f$ may be defined at $x=0$ and be continuous.
My procedure:
(1) $$\lim_{x\to 0} \frac{arctan(9bx)}{x} = \lim_{x\to 0} \frac{arctan(9bx)-arctan(9b*0)}{x} = \frac d{dx}arctan(9bx)|_{x=0}=\Bigl(\frac{1}{1+(9bx)^2}9b\Bigr)|_{x=0}=9b=\lim_{x\to 0^{+}} \frac{arctan(9bx)}{x}=\lim_{x\to 0^{-}} \frac{arctan(9bx)}{x}$$
Then the limit $\lim_{x\to 0^{-}} \frac{arctan(9bx)}{x}$ exist.
(2) $$\lim_{x\to 0^{+}} (1 + 2^{\frac{3}{x}})^{bsin(x)} = \infty^0 \;(indetermination)$$ The thing is I don´t really know how to calculate the second limit. Any hint in how to proceed with the limit?. Preferably without using L'Hopitals rule.
The problem is to compute the right-side limit. Assume henceforth $x>0$. $$\log(1+2^{3/x})^{b\sin x}=b\sin x\log(1+2^{3/x})=b\sin x\log(2^{3/x}(1+2^{-3/x}))=b\sin x\frac{3}{x}\log 2+o(x)$$ So the logarithm of the expression tends, as $x\downarrow 0$, to $3b\log 2$.