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.
Best Answer
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?