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?

I'll use the famous limit
$$\left(1+\frac{a}{x+1}\right)^x\approx\left(1+\frac{a}{x}\right)^x\to e^a$$
We have
$$x \ln \frac{x+1}{x+10}=x \ln \frac{x+1}{x+1+9}=-x\ln\left( 1+\frac{9}{x+1} \right)=-\ln\left( 1+\frac{9}{x+1} \right)^x\to-9$$
