Evalutating $\underset{x\to \mathrm{\infty}}{lim}(x+1{)}^{k}-(x{)}^{k}$, given $0<k<1$

2d3vljtq
2022-07-10
Answered

Evalutating $\underset{x\to \mathrm{\infty}}{lim}(x+1{)}^{k}-(x{)}^{k}$, given $0<k<1$

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haingear8v

Answered 2022-07-11
Author has **13** answers

You can just rewrite the expression and use L'Hôpital's rule:

$\underset{x\to +\mathrm{\infty}}{lim}((x+1{)}^{k}-{x}^{k})=\underset{x\to +\mathrm{\infty}}{lim}{\displaystyle \frac{(1+1/x{)}^{k}-1}{1/{x}^{k}}}=\underset{x\to +\mathrm{\infty}}{lim}{\displaystyle \frac{(-k/{x}^{2})(1+1/x{)}^{k-1}}{-k{x}^{-k-1}}}=\underset{x\to +\mathrm{\infty}}{lim}(x+1{)}^{k-1}=0$

$\underset{x\to +\mathrm{\infty}}{lim}((x+1{)}^{k}-{x}^{k})=\underset{x\to +\mathrm{\infty}}{lim}{\displaystyle \frac{(1+1/x{)}^{k}-1}{1/{x}^{k}}}=\underset{x\to +\mathrm{\infty}}{lim}{\displaystyle \frac{(-k/{x}^{2})(1+1/x{)}^{k-1}}{-k{x}^{-k-1}}}=\underset{x\to +\mathrm{\infty}}{lim}(x+1{)}^{k-1}=0$

civilnogwu

Answered 2022-07-12
Author has **6** answers

Use Taylor polynomial: as $h\to 0$, we have $(1+h{)}^{k}=1+kh+o(h)$

As $x\to \mathrm{\infty}$

$(x+1{)}^{k}={x}^{k}{(1+\frac{1}{x})}^{k}={x}^{k}(1+\frac{k}{x}+o(1/x))={x}^{k}+k{x}^{k-1}+o({x}^{k-1})\phantom{\rule{0ex}{0ex}}(x+1{)}^{k}-{x}^{k}=k{x}^{k-1}+o({x}^{k-1})$

Now $k<1$ so $k-1<0$ and thus $\underset{x\to \mathrm{\infty}}{lim}{x}^{k-1}=0$. Conclude

$(x+1{)}^{k}-{x}^{k}\to 0\phantom{\rule{1em}{0ex}}\text{as}x\to \mathrm{\infty}$

or more pricisely

$(x+1{)}^{k}-{x}^{k}\sim k{x}^{k-1}\phantom{\rule{1em}{0ex}}\text{as}x\to \mathrm{\infty}$

Example:

$\sqrt{x+1}-\sqrt{x}\sim \frac{1}{2\sqrt{x}}\phantom{\rule{1em}{0ex}}\text{as}x\to \mathrm{\infty}$

As $x\to \mathrm{\infty}$

$(x+1{)}^{k}={x}^{k}{(1+\frac{1}{x})}^{k}={x}^{k}(1+\frac{k}{x}+o(1/x))={x}^{k}+k{x}^{k-1}+o({x}^{k-1})\phantom{\rule{0ex}{0ex}}(x+1{)}^{k}-{x}^{k}=k{x}^{k-1}+o({x}^{k-1})$

Now $k<1$ so $k-1<0$ and thus $\underset{x\to \mathrm{\infty}}{lim}{x}^{k-1}=0$. Conclude

$(x+1{)}^{k}-{x}^{k}\to 0\phantom{\rule{1em}{0ex}}\text{as}x\to \mathrm{\infty}$

or more pricisely

$(x+1{)}^{k}-{x}^{k}\sim k{x}^{k-1}\phantom{\rule{1em}{0ex}}\text{as}x\to \mathrm{\infty}$

Example:

$\sqrt{x+1}-\sqrt{x}\sim \frac{1}{2\sqrt{x}}\phantom{\rule{1em}{0ex}}\text{as}x\to \mathrm{\infty}$

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