How to solve this limit problem? $\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )$ 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.
 A: $$\lim _{ n\to \infty  } \left( \sqrt { n+\sqrt { n }  } -\sqrt { n-\sqrt { n }  }  \right) =\lim _{ n\to \infty  } \frac { \left( \sqrt { n+\sqrt { n }  } -\sqrt { n-\sqrt { n }  }  \right) \left( \sqrt { n+\sqrt { n }  } +\sqrt { n-\sqrt { n }  }  \right)  }{ \left( \sqrt { n+\sqrt { n }  } +\sqrt { n-\sqrt { n }  }  \right)  }\\\\ =\lim _{ n\to \infty  } \frac { 2\sqrt { n }  }{ \left( \sqrt { n+\sqrt { n }  } +\sqrt { n-\sqrt { n }  }  \right)  } =\lim _{ n\to \infty  } \frac { 2\sqrt { n }  }{ \sqrt { n } \left( \sqrt { 1+\frac { 1 }{ \sqrt { n }  }  } +\sqrt { 1-\frac { 1 }{ \sqrt { n }  }  }  \right)  } =\\=\lim _{ n\to \infty  } \frac { 2 }{ \left( \sqrt { 1+\frac { 1 }{ \sqrt { n }  }  } +\sqrt { 1-\frac { 1 }{ \sqrt { n }  }  }  \right)  } =1$$
A: 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.
A: 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  $1$.
A: 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 $f(0)=1$.
A: 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 $$A_{100}=\frac{1281607}{1280000}\approx 1.0012554688$$$
