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

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.

One Answer

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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