<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] $$\begin{aligned}\frac{1}{\ln(x+1)-\ln(x)}-x 
&= \frac{1}{\ln\left(1 + \frac{1}{x}\right)}-x \\
&= \frac{1}{\frac{1}{x} - \frac{1}{2x^2}} -x + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= \frac{1}{2} + o(1) \\
 \end{aligned}$$
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