Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logarithm laws, we have
$$\lim_{x\rightarrow \infty} \left(\dfrac{1}{\ln\frac{x+1}{x}}-x\right).$$
Simplyfing further, we have:
$$\lim_{x\rightarrow \infty} \dfrac{1-x\ln\frac{x+1}{x}}{\ln\dfrac{x+1}{x}}= \lim_{x\rightarrow \infty} \frac{1-\ln(1+\frac{1}{x})^x}{\ln(1+\frac{1}{x})} = \frac{1-e}{0} \rightarrow -\infty.$$
However, the answer sheet tells me that it's $1/2$, and I don't really see where I did something wrong in the solution. Thanks.

\begin{gather*}
\lim _{x\rightarrow \infty }\frac{1}{\ln( 1+x) -\ln x} -x\\
=\lim _{x\rightarrow \infty }\frac{1}{\ln\left( 1+\frac{1}{x}\right)} -x\\
=\lim _{x\rightarrow \infty }\frac{1-x\cdotp \ln\left( 1+\frac{1}{x}\right)}{\ln\left( 1+\frac{1}{x}\right)}\\
Let\ us\ make\ a\ substitution\ h=\frac{1}{x}\\
Then,\ h\rightarrow 0\ as\ x\rightarrow \infty \\
=\lim _{h\rightarrow 0}\frac{1-\frac{1}{h} \cdotp \ln( 1+h)}{\ln( 1+h)}\\
We\ know\ that\ \\
\ln( 1+h) =h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...\\
Hence,\ \\
\lim _{h\rightarrow 0}\frac{1-\frac{1}{h} \cdotp \ln( 1+h)}{\ln( 1+h)}\\
=\lim _{h\rightarrow 0}\frac{1-\frac{1}{h} \cdotp \left( h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...\right)}{h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...}\\
=\lim _{h\rightarrow 0}\frac{1-1+\frac{h}{2} -\frac{h^{2}}{3} +\frac{h^{3}}{4} -...}{h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...}\\
=\lim _{h\rightarrow 0}\frac{\frac{h}{2} -\frac{h^{2}}{3} +\frac{h^{3}}{4} -...}{h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...}
\end{gather*}
Can you take it from here? Good luck!
