<s>[INST] 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.
 [/INST] I shall try to evaluate the limit by using L’Hospital Rule twice.
$$
\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow \infty}\left(\frac{1}{\ln (x+1)-\ln x}-x\right)\\
\begin{aligned}=& \lim _{x \rightarrow \infty} \frac{1-x \ln \left(\frac{x+1}{x}\right)}{\ln \left(\frac{x+1}{x}\right)} \quad\left(\frac{0}{0}\right) \\=& \lim _{x \rightarrow \infty} \frac{-x\left(\frac{x}{x+1}\right)\left(-\frac{1}{x^{2}}\right)-\ln \left(\frac{x+1}{x}\right)}{\frac{x}{x+1}\left(-\frac{1}{x^{2}}\right)} \\
=& -\lim _{x \rightarrow \infty} \frac{\frac{1}{x+1}-\ln \left(\frac{x+1}{x}\right)}{\frac{1}{x}-\frac{1}{x+1}} \quad\left(\frac{0}{0}\right) \\=& -\lim _{x \rightarrow \infty} \frac{-\frac{1}{(x+1)^{2}}-\frac{x}{x+1}\left(-\frac{1}{x^{2}}\right)}{-\frac{1}{x^{2}}+\frac{1}{(x+1)^{2}}} 
\end{aligned}\end{array}\\ \begin{array}{l}
\end{array}
$$
Simplifying the quotient yields
$$
\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow \infty}\left(\frac{1}{\ln (x+1)-\ln x}-x\right) \\ \displaystyle =\lim _{x \rightarrow \infty} \frac{\frac{1}{(x+1)^{2}}-\frac{1}{x(x+1)}}{-\frac{1}{x^{2}}+\frac{1}{(x+1)^{2}}} \\
=\displaystyle \lim _{x \rightarrow \infty} \frac{x^{2}-x(x+1)}{-(x+1)^{2}+x^{2}} \\
=\displaystyle \lim _{x \rightarrow \infty} \frac{-x}{-2 x+1} \\
=\dfrac{-1}{-2+\frac{1}{x}} \\
=\dfrac{1}{2}
\end{array}
$$
</s>