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I know that $\lim_{n\to ∞}$ $\frac {\sqrt{n} + \sqrt[3]{n} + \sqrt[4]{n}}{\sqrt{2 n + 1}} = \frac {1}{\sqrt{2}}$

but i don't really follow the steps in between.

Anyone keen on helping out? Any help would be greatly appreciated.

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    "but i don't really follow the steps in between." Which steps? Are we supposed to guess them? – 2017-02-12
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    Divide top and bottom of the LHS by $\sqrt{n}$, – 2017-02-12
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    The steps needed to arrive to the answer. Some hints, if anybody had a clue. Sorry for not being clear. – 2017-02-12

4 Answers 4

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\begin{align*} \frac {\sqrt{n} + \sqrt[3]{n} + \sqrt[4]{n}}{\sqrt{2 n + 1}} &= \sqrt\frac {n} {2n+1} + \sqrt\frac {n^{2/3}}{2n+1} + \sqrt\frac {n^{1/2}}{2n+1}\\ &= \sqrt\frac {1}{2+\frac{1}{n}} + \sqrt\frac {\frac{1}{n^{1/3}}}{2+\frac{1}{n}} + \sqrt\frac {\frac{1}{n^{1/2}}}{2+\frac{1}{n}} \end{align*}

which leads to, since $n \mapsto n^{q}$ is continuos for all $q > 0\colon$

\begin{align*} \lim_{n\rightarrow\infty}\frac {\sqrt{n} + \sqrt[3]{n} + \sqrt[4]{n}}{\sqrt{2 n + 1}} &= \sqrt{\lim_{n\rightarrow\infty}\frac {1}{2+\frac{1}{n}}} + \sqrt{\lim_{n\rightarrow\infty}\frac {\frac{1}{n^{1/3}}}{2+\frac{1}{n}}} + \sqrt{\lim_{n\rightarrow\infty}\frac {\frac{1}{n^{1/2}}}{2+\frac{1}{n}}}\\ &= \sqrt{\frac 1 2}\; = \frac{1}{\sqrt{2}}. \end{align*}

3

Put $n=m^{12}$ and compute

$$\lim_{m\to +\infty}\frac{m^{6}+m^4+m^3}{m^6 \sqrt{2} }\sqrt{\frac{ 2 }{ 2+m^{-12} }}$$

which gives $\frac{1}{\sqrt{2}}$.

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    Also, this isn't really much of a "hint" as it is a full solution, – 2017-02-12
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    i made the correction – 2017-02-12
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    My mistake on the rollback, although ugly, you are correct to have included that squareroot on the side to account for the fact that it was a $\sqrt{2n+1}$ instead of a $\sqrt{2n}$ – 2017-02-12
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    That is pretty elegant. – 2017-02-12
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    @Obriareos Thanks a lot for your comment. – 2017-02-12
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Use equivalents:

$\sqrt n+\sqrt[3]n+\sqrt[4]n\sim_\infty \sqrt n$, so $$\frac{\sqrt n+\sqrt[3]n+\sqrt[4]n}{\sqrt{2n}}\sim_\infty \frac{\sqrt n}{\sqrt{2n}}=\frac1{\sqrt2}.$$

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    Very good style. When it comes to the asymptotic behaviour the Landau notation (or asymptotic equivalence) should always be used. – 2017-02-12
1

The standard (?) technique with rational functions involving radicals and/or polynomials for limits to infinity: Divide the top and bottom by something simple:

$$\begin{align}\frac{\sqrt x+\sqrt[3]x+\sqrt[4]x}{\sqrt{2x+1}}&=\frac{x^{1/2}+x^{1/3}+x^{1/4}}{(2x+1)^{1/2}}\\&=\frac{1+\frac1{x^{1/6}}+\frac1{x^{1/4}}}{(2+\frac1x)^{1/2}}\\&\to\frac{1+0+0}{(2+0)^{1/2}}\\&=\frac1{\sqrt2}\end{align}$$