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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$

Could someone help me through this problem? Calculate $\displaystyle\lim_{n \to{+}\infty}{(\sqrt{n^{2}+n}-n)}$


3 Answers 3


We have:

$$\sqrt{n^{2}+n}-n=\frac{(\sqrt{n^{2}+n}-n)(\sqrt{n^{2}+n}+n)}{\sqrt{n^{2}+n}+n}=\frac{n}{\sqrt{n^{2}+n}+n}$$ Therefore:


And since: $\lim\limits_{n\to +\infty}\frac{1}{n}=0$

It follows that:

$$\boxed{\,\,\lim\limits_{n\to +\infty}(\sqrt{n^{2}+n}-n)=\dfrac{1}{2}\,\,}$$


Guide: Rationalize,

$$\left(\sqrt{n^2+n}-\sqrt{n^2}\right)\cdot \frac{\sqrt{n^2+n}+\sqrt{n^2}}{\sqrt{n^2+n}+\sqrt{n^2}}=\frac{n}{\sqrt{n^2+n}+\sqrt{n^2}}$$

Now divide numerator and denominator by $n$. Remember $\frac{1}{n}\sqrt{\square}=\sqrt{\frac{1}{n^2}\square}$.


Here's an answer that is probably not within the intended scope but it's nice anyway...

Let $x=1/n$. Then $$ \lim_{n\to{+}\infty}{\sqrt{n^{2}+n}-n} = \lim_{x\to0}{\sqrt{\frac1{x^2}+\frac1x}-\frac1x} = \lim_{x\to0}{\sqrt{\frac{1+x}{x^2}}-\frac1x} = \lim_{x\to0}{\frac{\sqrt{1+x}}{x}-\frac1x}= \lim_{x\to0}{\frac{\sqrt{1+x}-1}{x-0}} = f'(0) = \frac12 $$ for $f(x)=\sqrt{1+x}$.

(There's a small technicality that actually $x\to0^+$ but let's overlook that.)

  • 1
    $\begingroup$ You can just take $x \rightarrow 0^+$ the whole way through and conclude as you do, since if $\lim_{x \rightarrow 0} g(x)$ exists, then so does $\lim_{x \rightarrow 0^+} g(x)$ and the two limits are equal. (Minor point: you have a $1/n$ that should be a $1/x$.) $\endgroup$ Apr 24, 2012 at 23:46
  • $\begingroup$ @MichaelJoyce, that's the small technicality I meant. And thanks for finding that typo. $\endgroup$
    – lhf
    Apr 24, 2012 at 23:50
  • 1
    $\begingroup$ $x\to 0^+$ is used when you wrote $\sqrt{\frac{1+x}{x^2}}=\frac{\sqrt{1+x}}{x}$ because $\sqrt{x^2}=|x|$. $\endgroup$
    – user236182
    Nov 2, 2017 at 20:04

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