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$$ \lim_{n\to \infty}\left(\frac{1}{n^4}{+3^{\frac{2}{2+n}}}\right)^{n} $$
So i re-write it like:
$\lim_{n\to \infty}e^{n\ln{\frac{1}{n^4}\ln3^{\frac{2}{2+n}}}}$ $=$ $e^{\frac{2n}{2+n}\ln{\frac{1}{n^4}\ln3}}=e^{{2}\ln{\frac{1}{n^4}\ln3}}$

So here, $\ln{\frac{1}{n^4}}$ give us minus infinity, but i think that somewhere i lost a way, so i need some hints

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    It looks like you incorrectly assumed the log of a sum is the product of the logs. – 2017-01-09
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    @GEdgar, i find it thx – 2017-01-09

3 Answers 3

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Put $n=\frac{1}{y}$ .The limit reduces to

$L=\lim_{y\to 0}\left(y^4+3^{\frac{2y}{2y+1}}\right)^{\frac{1}{y}}$

Take log on both sides

$\log_{e}L=\lim_{y\to 0} \log_{e}\left(y^4+3^{\frac{2y}{2y+1}}\right)^{\frac{1}{y}}$

$\log_{e}L=\lim_{y\to 0}\frac{1}{y} \log_{e}\left(y^4+3^{\frac{2y}{2y+1}}\right)$

$\log_{e}L=\lim_{y\to 0}\frac{ \log_{e}\left(1+y^4+3^{\frac{2y}{2y+1}}-1\right)}{y^4+3^{\frac{2y}{2y+1}}-1}\frac{y^4+3^{\frac{2y}{2y+1}}-1}{y}$

$\log_{e}L=\lim_{y\to 0}\frac{ \log_{e}\left(1+y^4+3^{\frac{2y}{2y+1}}-1\right)}{y^4+3^{\frac{2y}{2y+1}}-1}\left(y^3+\frac{3^{\frac{2y}{2y+1}}-1}{\frac{2y}{2y+1}}\frac{2}{2y+1}\right)=1\left(2\log_{e}3\right)=\log_{e}9$

As $y\to 0$ it is obvious that $y^4+3^{\frac{2y}{2y+1}}-1\to 0$.

So here the standard result $\lim_{x\to 0}\frac{\log_{e}(1+x)}{x}=1$ has been used.

Also as $y\to 0$ it is obvious that $\frac{2y}{2y+1}\to 0$

so the standard limit $\lim_{x\to 0}\frac{a^x-1}{x}=\log_{e}a$ can be used.

Hence,$L=9$

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    Can u show more information how did u get 2*ln3 – 2017-01-09
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    I have edited my answer – 2017-01-09
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Note that $$ \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a} {n}} \right)^{\,n} = e^{\,a} \quad \quad \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{f(n)}} {n}} \right)^{\,n} \ne \mathop {\lim }\limits_{n\; \to \;\infty } e^{\,f(n)} $$ that's the fault in your derivation. Consider instead that we have: $$ \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a} {{\left( {n + \alpha } \right)}} + \frac{b} {{\left( {n + \beta } \right)^{\,2} }} + \cdots } \right)^{\,n + \gamma } = e^{\,a} $$

Therefore we shall try and expand $3^{2/(2+n)}$ in powers of the exponent, i.e. Taylor of $3^x \quad | \, x=2/(2+n) \approx 0$ $$ \begin{gathered} \mathop {\lim }\limits_{n\; \to \;\infty } \left( {\frac{1} {{n^{\,4} }} + 3^{\,\frac{2} {{2 + n}}} } \right)^{\,n} = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {e^{\frac{{2\ln 3}} {{2 + n}}} + \frac{1} {{n^{\,4} }}} \right)^{\,n} = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{2\ln 3}} {{2 + n}} + \frac{{\left( {2\ln 3} \right)^2 }} {2}\frac{1} {{\left( {2 + n} \right)^2 }} + \cdots + \frac{1} {{n^{\,4} }}} \right)^{\,n} = \hfill \\ = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{2\ln 3}} {{2 + n}}} \right)^{\,n} = e^{2\ln 3} = 9 \hfill \\ \end{gathered} $$

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    U have used Taylor expansion? I didnt understand why u plus and minus 2 in last 1/(2+n-2) is it should be just look like last term , and about that -2 in power u canceled by adding +2 to the n in the n→inf? Is it ok? Sorr if the question kind of stupid – 2017-01-09
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    @МузаффарШакаров Well, I tried and make more readable my answer. It should solve now your questions, if not do not hesitate to ask again. – 2017-01-09
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We can see that it is " tending to one" raised to the power "tending to infinity" then we can use this as a shortcut

Then use L'Hospital rule to simplify