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.
 A: First, we have
$$\sqrt[10]{n^{10}+8n^9}-n=e^{\log(\sqrt[10]{n^{10}+8n^{9}}) }-n\ne e^{\log(\sqrt[10]{n^{10}+8n^{9}}-n) }$$
Second, $\lim_{n\to\infty}\frac{e^{\alpha_n}-1}{\alpha_n}=1$ and $\lim_{n\to\infty}\frac{\log(1+\beta_n)}{\beta_n}=1$ for $\alpha_n\to 0$ and $\beta_n\to 0$, respectively.
So, how shall we proceed?
Let's first write $\sqrt[10]{n^{10}+8n^9}-n$ as
$$\begin{align}
\sqrt[10]{n^{10}+8n^9}-n&=n\left(\sqrt[10]{1+\frac8n}-1\right)\\\\
&=n\left(e^{\frac1{10}\log\left(1+\frac8n\right)}-1\right)\\\\
&=n\left(\frac{e^{\frac1{10}\log\left(1+\frac8n\right)}-1}{\frac1{10}\log\left(1+\frac8n\right)}\right)\times \left(\frac{\frac1{10}\log\left(1+\frac8n\right)}{\frac8n}\right)\times\left(\frac8n\right)\\\\
\end{align}$$
Can you finish now?
A: $$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n)=\lim\limits_{n \rightarrow \infty}n \frac{\sqrt[10]{1+\frac{8}{n}}-1}{\frac{8}{n}}\cdot\frac{8}{n}=\frac{8}{10}$$
Using, for $x \to 0$
$$\frac{e^{\alpha \ln (1+x)}-1}{\alpha \ln (1+x)}\cdot \frac{\alpha \ln (1+x)}{x}=\frac{e^{\alpha \ln (1+x)}-1}{x}=\frac{(1+x)^\alpha-1}{x}\to \alpha$$
A: $$(\sqrt[10]{n^{10} + 8n^9} - n)=n\left(\sqrt[10]{1+\frac{8}{n}}-1\right) $$
$$\sqrt[10]{1+\frac{8}{n}}=\sum_{k=0}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}=1+\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}$$
$$n\left(\sqrt[10]{1+\frac{8}{n}}-1\right)=\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^{k-1}}=\frac{4}{5}-\frac{72}{25 n}+\cdots $$
