Prove that exist $b gt 0$, so that $f$ may be defined at $x=0$ and be continuous.

Given the function $$
f(x) =
begin{cases}
(1 + 2^{frac{3}{x}})^{bsin(x)} &quad if quad xgt 0 \
\
frac{arctan(9bx)}{x} &quad if quad xlt 0 \
end{cases}
$$

Prove that exist $b gt 0$, so that $f$ may be defined at $x=0$ and be continuous.

My procedure:

(1) $$lim_{xto 0} frac{arctan(9bx)}{x} = lim_{xto 0} frac{arctan(9bx)-arctan(9b*0)}{x} = frac d{dx}arctan(9bx)|_{x=0}=Bigl(frac{1}{1+(9bx)^2}9bBigr)|_{x=0}=9b=lim_{xto 0^{+}} frac{arctan(9bx)}{x}=lim_{xto 0^{-}} frac{arctan(9bx)}{x}$$

Then the limit $lim_{xto 0^{-}} frac{arctan(9bx)}{x}$ exist.

(2) $$lim_{xto 0^{+}} (1 + 2^{frac{3}{x}})^{bsin(x)} = infty^0 ;(indetermination)$$
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.