Limit of $\sqrt[10]{n^{10} + 8n^9} - n$ as $n \rightarrow \infty$ using two "standard" limits

Question

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.

Answer 1

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?

Answer 2

$$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n)=\lim\limits_{n \rightarrow \infty}n \frac{\sqrt[10]{1+\frac{8}{n}}-1}{\frac{8}{n}}\cdot\frac{8}{n}=\frac{8}{10}$$ Using, for $x \to 0$ $$\frac{e^{\alpha \ln (1+x)}-1}{\alpha \ln (1+x)}\cdot \frac{\alpha \ln (1+x)}{x}=\frac{e^{\alpha \ln (1+x)}-1}{x}=\frac{(1+x)^\alpha-1}{x}\to \alpha$$

Answer 3

$$(\sqrt[10]{n^{10} + 8n^9} - n)=n\left(\sqrt[10]{1+\frac{8}{n}}-1\right) $$ $$\sqrt[10]{1+\frac{8}{n}}=\sum_{k=0}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}=1+\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}$$ $$n\left(\sqrt[10]{1+\frac{8}{n}}-1\right)=\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^{k-1}}=\frac{4}{5}-\frac{72}{25 n}+\cdots $$