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?
 A: $$
\begin{align}
\lim_{x\to\infty}x\log\left(\frac{x+1}{x+10}\right)
&=\lim_{x\to\infty}x\log\left(1-\frac{9}{x+10}\right)\\
&=\lim_{x\to\infty}\frac{\log\left(1-\frac{9}{x+10}\right)}{-\frac{9}{x+10}}\left(-\frac{9x}{x+10}\right)\\[9pt]
&=\lim_{u\to0}\frac{\log(1+u)}{u}\lim_{x\to\infty}\left(-\frac{9x}{x+10}\right)\\[9pt]
&=1\cdot(-9)\\[18pt]
&=-9
\end{align}
$$
A: Putting $h=\frac1x$ and assuming Natural Logarithm 
$$\lim_{x\to\infty}x\cdot \ln \frac{x+1}{x+10} =\lim_{h\to0}\frac{\ln\frac{1+h}{1+10h}}h$$
$$\text{As }\ln_e \frac ab=\ln_e a-\ln_e b, \ln\frac{1+h}{1+10h}=\ln(1+h)-\ln(1+10h)$$
$$\implies \lim_{h\to0}\frac{\ln\frac{1+h}{1+10h}}h=\lim_{h\to0}\frac{\ln(1+h)-\ln(1+10h)}h$$ 
$$=\lim_{h\to0}\frac{\ln(1+h)}h-10\cdot\lim_{h\to0}\frac{\ln(1+10h)}{10h} $$ 
Do you know, $$\lim_{x\to0}\frac{\ln(1+x)}x=??$$
A: 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
$$x \ln \frac{x+1}{x+10}=x \ln \frac{x+1}{x+1+9}=-x\ln\left( 1+\frac{9}{x+1} \right)=-\ln\left( 1+\frac{9}{x+1} \right)^x\to-9$$
A: 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).
