<s>[INST] Limit of $\sqrt[10]{n^{10} + 8n^9} - n$ as $n \rightarrow \infty$ using two "standard" limits I must use the two standard limits $\lim\limits_{n \rightarrow \infty} \frac{e^{\alpha_n}-1}{\alpha_n} = 1$ and $\lim\limits_{n \rightarrow \infty} \frac{\ln(1+\beta_n)}{\beta_n} = 1$ if  $\alpha_n, \beta_n \rightarrow 0$ so multiplying by conjugate approach won't work.
My attempt:
$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n) = \lim\limits_{n \rightarrow \infty} e^{\ln(\sqrt[10]{n^{10} + 8n^9} - n)} = e^{\lim\limits_{n \rightarrow \infty} (\ln(\sqrt[10]{n^{10} + 8n^9} - n))}$
I am not sure how to proceed further. Any help/hints will be appreciated.
 [/INST] $$(\sqrt[10]{n^{10} + 8n^9} - n)=n\left(\sqrt[10]{1+\frac{8}{n}}-1\right) $$
$$\sqrt[10]{1+\frac{8}{n}}=\sum_{k=0}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}=1+\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}$$
$$n\left(\sqrt[10]{1+\frac{8}{n}}-1\right)=\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^{k-1}}=\frac{4}{5}-\frac{72}{25 n}+\cdots $$
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