<s>[INST] Find $\lim \limits_{n \to \infty}(n^5+4n^3)^{1/5}-n$ 
$$\lim \limits_{n \to \infty}(n^5+4n^3)^{1/5}-n=?$$ 

I see that   $$\lim_{n \to \infty}(n^5+4n^3)^{1/5}-n=\lim_{n \to \infty}n[(1+ \frac {4}{n^2})^{1/5}-1]=\lim_{z \to 0} \frac {1}{z}[(1+ {4}{z^2})^{1/5}-1]$$,where $n=\frac {1}{z}$. Now I do not know how to proceed.
Can someone point me in the right direction? Thanks in advance for your time.
 [/INST] Two methods:


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*Clever, but requiring to be clever: conjugate quantities. 


Here, the identity $(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)=a^5-b^5$ with $a=(n^5+4n^3)^{1/5}$ and $b=n$ yields $a^5-b^5=4n^3$. Furthermore, $a\sim b\sim n$ hence each term in the sum $a^4+a^3b+a^2b^2+ab^3+b^4$ is equivalent to $n^4$. Can you finish?


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*Not clever, but sure to win: limited expansions.


Here, $(1+x)^{1/5}=1+\frac15x+o(x)$ when $x\to0$ hence $(1+4/n^2)^{1/5}=1+O(1/n^2)$. Furthermore, $(n^5+4n^3)^{1/5}=n(1+4/n^2)^{1/5}$. Can you finish?
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