<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] If we are allowed to use series expansions, then the problem is fairly trivial. We want to evaluate $$\lim_{x\to \infty} \frac{1 - x \log \left(1+ \frac1x\right)}{\log \left(1+ \frac1x\right)}$$
The series expansion of $\log \left(1+ \frac1x\right)$ is given by $$\log \left(1+ \frac1x\right) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \frac{1}{4x^4} + \ldots$$
Plugging this in, we find that $$\frac{1 - x \log \left(1+ \frac1x\right)}{\log \left(1+ \frac1x\right)} = \frac{1 - x\left(\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} + \ldots \right)}{\left(\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} + \ldots\right) } = \frac{\left(\frac{1}{2} - \frac{1}{3x} + \frac{1}{4x^2} \ldots \right)}{\left(1 - \frac{1}{2x} + \frac{1}{3x^2} + \ldots \right)} \xrightarrow{x\to\infty} \frac12$$
Thus, $$\boxed{\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x = \frac12}$$
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