Find the limit of $a_n$ = $(n+\sqrt n)^\frac{1}{4} - n^\frac{1}{4}$. Find the limit of $a_n = (n+\sqrt n)^\frac{1}{4} - n^\frac{1}{4}$.
I know that it converges and my first thought was to multiply by the conjugate, but then I had square roots at the top and fourth roots at the bottom, unless I missed a factorisation I wasn’t sure how to use the algebra of limits from there to proceed.
I was also thinking of squeeze theorem with 0 as the lower bound but I couldn’t find an upper bound (maybe $\frac{1}{n}$) for large values of $n$, but that would have been a guess. I guess instead I could use the monotone convergence theorem but I’m sure I’m missing something obvious.
I just need this answering - Finally, I was considering showing that this sequence $\leq n$, and then using epsilon-delta. I guess that would work , no?!
 A: The usual techniques, in case of an indeterminate form $\infty-\infty$, is to isolate the common factor (of the two terms of the difference) which tends to $\infty$, like this:$$\begin{align}a_n&=n^\frac14\left(\left(1+\frac1{\sqrt n}\right)^\frac14-1\right)\\&\sim n^\frac14\frac1{4\sqrt n}=\frac1{4\sqrt[4]n}\to0,
\end{align}$$
where the Taylor expansion $(1+h)^\alpha=1+\alpha h+o(h)$ has been applied to $h=\frac1{\sqrt n}$ and $\alpha=\frac14.$
But you could also have iterated your idea to multiply by the conjugate:
$$\begin{align}a_n&=\frac{(n+\sqrt n)^\frac12-n^\frac12}{(n+\sqrt n)^\frac14+n^\frac14}\\&=\frac{\sqrt n}{\left((n+\sqrt n)^\frac14+n^\frac14\right)\left((n+\sqrt n)^\frac12+n^\frac12\right)}\\&<\frac{\sqrt n}{\left(2n^\frac14\right)\left(2n^\frac12\right)}\\&=\frac1{4\sqrt[4]n}.
\end{align}$$
Since $a_n\ge0,$ the squeeze theorem then permits to conclude as before: $a_n\to0.$
Edit: as mentionned by robjohn in his (auto-deleted) answer, another way to prove that $a_n\le\frac1{4\sqrt[4]n}$ is to apply Bernoulli's inequality.
A: Your idea about multiplying by the conjugate works well. For $x>y>0$ we have
$$0<x-y=\frac{x^4-y^4}{x^3+x^2y+xy^2+y^3}\leq\frac{x^4-y^4}{y^3}.$$
Hence
$$0<a_n=(n+\sqrt n)^\frac{1}{4} - n^\frac{1}{4}\leq \frac{(n+\sqrt n)-n}{n^{3/4}}=n^{-\frac14}.$$
Now squeeze theorem gives the limit.
A: I'll show maybe not the most common method (since I believe that Adam Rubinson & Anne Bauval's methods are more common among students) but a trick that is worth knowing. It may be considered a some sort of generalisation of Feng's answer. So let us consider natural number $n\in\mathbb{N}$ and a function $f_n$ defined as follows $$[n,n+\sqrt{n}]\ni x \mapsto \sqrt[4]{x}\in\mathbb{R}. $$ Since the function $f_n$ fulfill the assumption of  Mean value theorem ther is a number $\xi_n\in(n,n+\sqrt{n})$ such that  $$\frac{f_n(n+\sqrt{n})-f_n(n)}{\sqrt{n}}=f'_n(\xi_n)$$ i.e. $$\sqrt[4]{n+\sqrt{n}}- \sqrt[4]{n} = \frac{\sqrt{n}}{4\sqrt[4]{\xi_n^3}}.$$ However, since we know the boundary of $\xi_n\in(n,n+\sqrt{n})$ we can easily see that  $$\frac{\sqrt{n}}{4\sqrt[4]{(n+\sqrt{n})^3}} \le \sqrt[4]{n+\sqrt{n}}- \sqrt[4]{n} = \frac{\sqrt{n}}{4\sqrt[4]{\xi_n^3}}\le\frac{\sqrt{n}}{4\sqrt[4]{n^3}}. $$
Now the Squeeze Theorem work. Of course the lower bound is eaven too tight but it was given for free. Moreover now the $N-\epsilon$ argument is easy since for a given  $\epsilon>0$ by taking $N\ge (1/4\epsilon)^{4}$ we have $$\Big|\sqrt[4]{n+\sqrt{n}}- \sqrt[4]{n}\Big|\le \frac{\sqrt{n}}{4\sqrt[4]{n^3}} \le \epsilon $$ for all $n> N$. And lastly for students that do not like Mean value theorem. This method can be expressed in terms of integer $$f_n(n+\sqrt{n})-f_n(n)=\int_{n}^{n+\sqrt{n}}f'_n(x)\,\mathrm{d} x.$$ Now by similar estimation of the function $f_n'$ we end up with the same ineqalites.
A: Using difference of two squares twice, notice that
$$ \left[ \left( n+ \sqrt{n} \right)^{1/2} + n^{1/2} \right] \cdot \left[ \left( n+ \sqrt{n} \right)^{1/4} + n^{1/4} \right] \cdot \left[ \left( n+ \sqrt{n} \right)^{1/4} - n^{1/4} \right] = n + \sqrt{n} - n = \sqrt{n}. $$
Therefore,
$$ 0\leq\left( n+ \sqrt{n} \right)^{1/4} - n^{1/4} = \frac{ \sqrt{n} }{ \left[ \left( n+ \sqrt{n} \right)^{1/2} + n^{1/2} \right] \cdot \left[ \left( n+ \sqrt{n} \right)^{1/4} + n^{1/4} \right] } $$
$$ \leq \frac{ \sqrt{n} }{ 2\sqrt{n} \cdot 2\sqrt{\sqrt{n}} } = \frac{1}{4\sqrt{\sqrt{n}} } \overset{n\to\infty}{\longrightarrow} 0. $$
