Mathematics Asked by Karl on August 25, 2020

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.

The problem is to compute the right-side limit. Assume henceforth $x>0$. $$log(1+2^{3/x})^{bsin x}=bsin xlog(1+2^{3/x})=bsin xlog(2^{3/x}(1+2^{-3/x}))=bsin xfrac{3}{x}log 2+o(x)$$ So the logarithm of the expression tends, as $xdownarrow 0$, to $3blog 2$.

Correct answer by uniquesolution on August 25, 2020

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