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?

By the Lagrange mean value theorem, for $0<a\le b$, we have
$$
\frac{b-a}{b}\le\log(b)-\log(a)\le\frac{b-a}{a}
$$
So,
$$
L(x)=x\log\left(\frac{x+1}{x+10}\right)=-x\log\left(\frac{x+10}{x+1}\right)=
-x(\log(x+10)-\log(x+1))\\
\implies -9\frac{x}{x+10}\le L(x)\le -9\frac{x}{x+1}
$$
which shows $L(x)\stackrel{x \to \infty}{\to} -9$, by the squeeze principle (theorem).
