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Evalutating $\underset{x\to \mathrm{\infty }}{lim}\left(x+1{\right)}^{k}-\left(x{\right)}^{k}$, given $0
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haingear8v
You can just rewrite the expression and use L'Hôpital's rule:
$\underset{x\to +\mathrm{\infty }}{lim}\left(\left(x+1{\right)}^{k}-{x}^{k}\right)=\underset{x\to +\mathrm{\infty }}{lim}\frac{\left(1+1/x{\right)}^{k}-1}{1/{x}^{k}}=\underset{x\to +\mathrm{\infty }}{lim}\frac{\left(-k/{x}^{2}\right)\left(1+1/x{\right)}^{k-1}}{-k{x}^{-k-1}}=\underset{x\to +\mathrm{\infty }}{lim}\left(x+1{\right)}^{k-1}=0$
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civilnogwu
Use Taylor polynomial: as $h\to 0$, we have $\left(1+h{\right)}^{k}=1+kh+o\left(h\right)$
As $x\to \mathrm{\infty }$
$\left(x+1{\right)}^{k}={x}^{k}{\left(1+\frac{1}{x}\right)}^{k}={x}^{k}\left(1+\frac{k}{x}+o\left(1/x\right)\right)={x}^{k}+k{x}^{k-1}+o\left({x}^{k-1}\right)\phantom{\rule{0ex}{0ex}}\left(x+1{\right)}^{k}-{x}^{k}=k{x}^{k-1}+o\left({x}^{k-1}\right)$
Now $k<1$ so $k-1<0$ and thus $\underset{x\to \mathrm{\infty }}{lim}{x}^{k-1}=0$. Conclude

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