$\lim\limits_{x\to \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ Having a hard time solving $\lim\limits_{x\to \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$.
I tried using $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$ , to rewrite the denominator so i got 
$\lim\limits_{x\to \infty} \frac{1}{\ln \frac {x+1}{x}}-x$
But from there im unsure how to use a series to substitute the ln part. 
 A: You can use Taylor series: for $x\to\infty$
$$\frac{1}{\ln(1+\frac1x)}=\left(\frac1x-\frac1{2x^2}+O(1/x^3)\right)^{-1}=x\left(1+\frac1{2x}+O(1/x^2)\right)$$
so by simple algebra you get $\frac12$ the desired limit.
A: Alternatively, make the change: $\ln (x+1)-\ln x=y \Rightarrow x=\frac{1}{e^y-1}$. Then:
$$\lim_{y\to 0+}\left(\frac{1}{y}-\frac{1}{e^y-1}\right)=\lim_{y\to0+}\frac{e^y-1-y}{ye^y-y}\overbrace{=}^{L}\lim_{y\to0+}\frac{e^y-1}{e^y+ye^y-1}\overbrace{=}^{L}\lim_{y\to0+}\frac{e^y}{2e^y+ye^y}=\frac12.$$
A: Alternatively, you can also set $x=1/t$. The limit becomes (after little algebra):
$\lim_{t\to 0+}\frac{1}{ln(1+t)}-\frac{1}{t}$ which is still indeterminate. Performing the subtraction by making one term: $\lim_{t\to 0+}\frac{t-ln(1+t)}{tln(1+t)}$ which is a zero over zero situation on which you may apply L'Hospital's Rule.In fact, you will have to apply it twice, but this is just a little annoying algebra. After the second time, you end up with $\lim_{t\to 0+}\frac{1}{1+ln(1+t)+1}$ from which $1/2$ follows.
