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I am stuck with the following question from my homework:

$\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )$

Using wolfram alpha gives me 1 for the solution. However, I would like to know how you come up with this result. I hope someone can explain this to me.

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    $\begingroup$ Hint: Conjugation, or "rationalizing the numerator" should do it. $\endgroup$ Oct 23, 2016 at 15:50
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    $\begingroup$ Multiply by $\dfrac{\sqrt{n+\sqrt n}+\sqrt{n-\sqrt n}}{\sqrt{n+\sqrt n}+\sqrt{n-\sqrt n}}$ and simplify. $\endgroup$ Oct 23, 2016 at 15:51
  • $\begingroup$ read all solutions $\endgroup$ Oct 23, 2016 at 16:00
  • $\begingroup$ Accept the solution that helps you most; and/or, if you have a question or two about one of the posts, ask it as a comment below the answer of interest; $\endgroup$
    – amWhy
    Oct 23, 2016 at 16:03
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    $\begingroup$ Duplicate: $\lim\limits_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$. (Found using Approach0.xyz) $\endgroup$
    – Workaholic
    Nov 1, 2016 at 21:25

5 Answers 5

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$$\lim _{ n\to \infty } \left( \sqrt { n+\sqrt { n } } -\sqrt { n-\sqrt { n } } \right) =\lim _{ n\to \infty } \frac { \left( \sqrt { n+\sqrt { n } } -\sqrt { n-\sqrt { n } } \right) \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) }{ \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) }\\\\ =\lim _{ n\to \infty } \frac { 2\sqrt { n } }{ \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) } =\lim _{ n\to \infty } \frac { 2\sqrt { n } }{ \sqrt { n } \left( \sqrt { 1+\frac { 1 }{ \sqrt { n } } } +\sqrt { 1-\frac { 1 }{ \sqrt { n } } } \right) } =\\=\lim _{ n\to \infty } \frac { 2 }{ \left( \sqrt { 1+\frac { 1 }{ \sqrt { n } } } +\sqrt { 1-\frac { 1 }{ \sqrt { n } } } \right) } =1$$

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  • $\begingroup$ I somehow struggle with the simple step of extracting $\sqrt{n}$ out of $\sqrt{n + \sqrt{n}}$. I come up with $\sqrt{n + \sqrt{n}} = \sqrt{n \left ( 1 + \frac{\sqrt{n}}{n} \right )} = \sqrt{n} \sqrt{ 1 + \frac{\sqrt{n}}{n}}$ $\endgroup$
    – Sam
    Oct 23, 2016 at 17:27
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    $\begingroup$ @Sam,note that $\frac { 1 }{ \sqrt { n } } =\frac { \sqrt { n } }{ \sqrt { n } \sqrt { n } } =\frac { \sqrt { n } }{ n } $ $\endgroup$
    – haqnatural
    Oct 23, 2016 at 17:29
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In these cases we use the following trick: multiply by $1=\dfrac{\sqrt{n+\sqrt{n}}+\sqrt{n+\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n+\sqrt{n}}}$.

You get $$\lim \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )\times \dfrac{\sqrt{n+\sqrt{n}}+\sqrt{n+\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n+\sqrt{n}}}=\lim \dfrac{n+\sqrt{n}-n+\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n+\sqrt{n}}}=\lim \dfrac{2\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n+\sqrt{n}}},$$

and I leave it to you from here.

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  • $\begingroup$ I hadn't seen them as I was writing my answer. No need to be rude! $\endgroup$
    – user194469
    Oct 23, 2016 at 15:55
  • $\begingroup$ I'm not being rude. I asked a question, and you responded with an understandable answer. I know how frustrating it can be when you respond relatively early, in answering a question, only to learn after posting, others have responded to. (I can relate!) Thanks for answering my question! $\endgroup$
    – amWhy
    Oct 23, 2016 at 15:56
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If you find $$ \lim_{x \to \infty} \bigl(\sqrt{x+\sqrt{x}} - \sqrt{x-\sqrt{x}}\bigr) $$ (the function rather than the sequence), then the sequence has the same limit.

Now try a good substitution, in this case $\sqrt{x}=1/t$, so we get $$ \lim_{t\to0^+}\frac{\sqrt{1+t}-\sqrt{1-t}}{t} $$ which coincides with the derivative at $0$ of $$ f(t)=\sqrt{1+t}-\sqrt{1-t} $$ provided it exists. Since $$ f'(t)=\frac{1}{2\sqrt{1+t}}+\frac{1}{2\sqrt{1-t}} $$ we have $f(0)=1$.

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Put $\sqrt{n}=x$

we want

$\lim_{x\to+\infty}x(\sqrt{1+\frac{1}{x}}-\sqrt{1-\frac{1}{x}})$.

but when $X\to+\infty$

$\sqrt{1+\frac{1}{X}}=1+\frac{1}{2X}(1+\epsilon(X))$

thus

our limit is $1$.

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Just another way.

Factor $\sqrt n$ in each radical. So $$A_n=\sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}}=\sqrt n \left(\sqrt{1+n^{-1/2} }-\sqrt{1-n^{-1/2} }\right)$$ Now, since $n$ is large, use $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ and replace $x$ by $\pm n^{-1/2}$ This would lead to $$A=\sqrt n \left(\left(1+\frac 12 n^{-1/2}-\frac 18 n^{-1}+\cdots\right)-\left(1-\frac 12 n^{-1/2}-\frac 18 n^{-1}+\cdots\right)\right)$$ $$A=\sqrt n \left(n^{-1/2}+\cdots\right)\approx 1$$

Adding more terms in the developements, you should arrive to

$$A_n=1+\frac{1}{8 n}+\frac{7}{128 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

For illustration, letus try using $n=100$; the exact value $$A_{100}=\sqrt{10} \left(\sqrt{11}-3\right)\approx 1.0012555012$$ while the above aproximation gives $$A_{100}=\frac{1281607}{1280000}\approx 1.0012554688$$$

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