# Solve this question \lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 +

Ali Marshall 2022-04-28 Answered
Solve this question
$\underset{x\to \mathrm{\infty }}{lim}{\left(x+3\right)}^{1+\frac{1}{x}}-{x}^{1+\frac{1}{x+3}}$
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Tyler Velasquez
Write the function of interest as $f\left(x\right)={×}^{\frac{1}{x}}g\left(x\right)$ with
$g\left(x\right)={\left(1+\frac{3}{x}\right)}^{1+\frac{1}{x}}-{x}^{\frac{1}{x+3}-\frac{1}{x}}={e}^{a\left(x\right)}-{e}^{b\left(x\right)}$
with
$a\left(x\right)=\left(1+\frac{1}{x}\right)\mathrm{log}\left(1+\frac{3}{x}\right)=\frac{3}{x}+o\left(\frac{1}{x}\right)$
and
$b\left(x\right)=-\frac{3\mathrm{log}\left(x\right)}{x\left(x+3\right)}=o\left(\frac{1}{x}\right)$
This yields ${e}^{a\left(x\right)}=1+\frac{3}{x}+o\left(\frac{1}{x}\right)$ and ${e}^{b\left(x\right)}=1+o\left(\frac{1}{x}\right)$ hence $g\left(x\right)=\frac{3}{x}+o\left(\frac{1}{x}\right)$
Likewise ${x}^{\frac{1}{x}}={e}^{\frac{\mathrm{log}\left(x\right)}{x}}=1+o\left(1\right)$ hence
$f\left(x\right)={×}^{\frac{1}{x}}g\left(x\right)=x\left(1+o\left(1\right)\right)\left(\frac{3}{x}+o\left(\frac{1}{x}\right)\right)=3+o\left(1\right)$
That is, f does converge and its limit is 3
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