<s>[INST] Proof that $ \lim_{x \to \infty} x \cdot \log(\frac{x+1}{x+10})$ is $-9$ Given this limit:  
$$ \lim_{x \to \infty} x \cdot \log\left(\frac{x+1}{x+10}\right) $$
I may use this trick:  
$$ \frac{x+1}{x+1} = \frac{x+1}{x} \cdot \frac{x}{x+10} $$
So I will have:  
$$ \lim_{x \to \infty} x \cdot \left(\log\left(\frac{x+1}{x}\right) + \log\left(\frac{x}{x+10}\right)\right)  = $$
$$ = 1 + \lim_{x \to \infty} x \cdot \log\left(\frac{x}{x+10}\right) $$
But from here I am lost, I still can't make it look like a fondamental limit. How to solve it?
 [/INST] $$
\begin{align}
\lim_{x\to\infty}x\log\left(\frac{x+1}{x+10}\right)
&=\lim_{x\to\infty}x\log\left(1-\frac{9}{x+10}\right)\\
&=\lim_{x\to\infty}\frac{\log\left(1-\frac{9}{x+10}\right)}{-\frac{9}{x+10}}\left(-\frac{9x}{x+10}\right)\\[9pt]
&=\lim_{u\to0}\frac{\log(1+u)}{u}\lim_{x\to\infty}\left(-\frac{9x}{x+10}\right)\\[9pt]
&=1\cdot(-9)\\[18pt]
&=-9
\end{align}
$$
</s>