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$\underset{x\to \mathrm{\infty}}{lim}{(x+3)}^{1+\frac{1}{x}}-{x}^{1+\frac{1}{x+3}}$

Ali Marshall
2022-04-28
Answered

Solve this question

$\underset{x\to \mathrm{\infty}}{lim}{(x+3)}^{1+\frac{1}{x}}-{x}^{1+\frac{1}{x+3}}$

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Tyler Velasquez

Answered 2022-04-29
Author has **19** answers

Write the function of interest as $f\left(x\right)={\times}^{\frac{1}{x}}g\left(x\right)$ with

$g\left(x\right)={(1+\frac{3}{x})}^{1+\frac{1}{x}}-{x}^{\frac{1}{x+3}-\frac{1}{x}}={e}^{a\left(x\right)}-{e}^{b\left(x\right)}$

with

$a\left(x\right)=(1+\frac{1}{x})\mathrm{log}(1+\frac{3}{x})=\frac{3}{x}+o\left(\frac{1}{x}\right)$

and

$b\left(x\right)=-\frac{3\mathrm{log}\left(x\right)}{x(x+3)}=o\left(\frac{1}{x}\right)$

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

$f\left(x\right)={\times}^{\frac{1}{x}}g\left(x\right)=x(1+o\left(1\right))(\frac{3}{x}+o\left(\frac{1}{x}\right))=3+o\left(1\right)$

That is, f does converge and its limit is 3

with

and

This yields

Likewise

That is, f does converge and its limit is 3

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The figure is something like:

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Antiderivative of $x{e}^{-c{x}^{2}}$

I need to define c in ${\int}_{0}^{\mathrm{\infty}}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).

Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.

Trying to use partial integration:

$\int f(x)g(x)dx=f(x)G(x)-\int {f}^{\prime}(x)G(X)$

and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:

${f}^{\prime}(x)=-2cxf(x)$

$G(X)=\frac{{x}^{2}}{2}$

${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-2cxf(x)\frac{{x}^{2}}{2}dx$

Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.

I need to define c in ${\int}_{0}^{\mathrm{\infty}}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).

Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.

Trying to use partial integration:

$\int f(x)g(x)dx=f(x)G(x)-\int {f}^{\prime}(x)G(X)$

and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:

${f}^{\prime}(x)=-2cxf(x)$

$G(X)=\frac{{x}^{2}}{2}$

${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-2cxf(x)\frac{{x}^{2}}{2}dx$

Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.

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