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.
 A: 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, $$\boxed{\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x = \frac12}$$
A: 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
$$
\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow \infty}\left(\frac{1}{\ln (x+1)-\ln x}-x\right) \\ \displaystyle =\lim _{x \rightarrow \infty} \frac{\frac{1}{(x+1)^{2}}-\frac{1}{x(x+1)}}{-\frac{1}{x^{2}}+\frac{1}{(x+1)^{2}}} \\
=\displaystyle \lim _{x \rightarrow \infty} \frac{x^{2}-x(x+1)}{-(x+1)^{2}+x^{2}} \\
=\displaystyle \lim _{x \rightarrow \infty} \frac{-x}{-2 x+1} \\
=\dfrac{-1}{-2+\frac{1}{x}} \\
=\dfrac{1}{2}
\end{array}
$$
A: \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!
A: $$\begin{aligned}\frac{1}{\ln(x+1)-\ln(x)}-x 
&= \frac{1}{\ln\left(1 + \frac{1}{x}\right)}-x \\
&= \frac{1}{\frac{1}{x} - \frac{1}{2x^2}} -x + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= \frac{1}{2} + o(1) \\
 \end{aligned}$$
A: 
Simplyfing further, we have: $$\lim_{x\to\infty}\frac{1−x\ln\left(\frac{x+1}{x}\right)}{\ln\left(\frac{x+1}{x}\right)}=\lim_{x\to\infty}\frac{1−\ln\left(1+\frac{1}{x}\right)^x}{\ln\left(1+\frac{1}{x}\right)}=\frac{1-e}{0}\rightarrow{-\infty}$$ 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.

You claimed $\lim_{x\to\infty}\ln\left[\left(1+\frac{1}{x}\right)^x\right]=e,$ which is wrong. You should have written $$\lim_{x\to\infty}\frac{1}{\ln(x+1)-\ln(x)}-x=\lim_{x\to\infty}\frac{1}{\ln\left(1+\frac{1}{x}\right)}-\frac{1}{\frac{1}{x}}$$ and substitute $y=\frac{1}{x}$, hence $$\lim_{x\to\infty}\frac{1}{\ln\left(1+\frac{1}{x}\right)}-\frac{1}{\frac{1}{x}}=\lim_{y\to0+}\frac{1}{\ln(1+y)}-\frac{1}{y}=\lim_{y\to0+}\frac{y-\ln(1+y)}{y\ln(1+y)}$$ Notice that for $y\gt0$, $y-\ln(1+y)\gt0$ and $y\ln(1+y)\gt0$ and $\lim_{y\to0+}y-\ln(1+y)=0$ and $\lim_{y\to0+}y\ln(1+y)=0$. Thus the conditions for L'Hospital's rule  are satisfied, and $$\lim_{y\to0+}\frac{y-\ln(1+y)}{y\ln(1+y)}=\lim_{y\to0+}\frac{1-\frac{1}{1+y}}{\ln(1+y)+\frac{y}{1+y}}=\lim_{y\to0+}\frac{y}{(1+y)\ln(1+y)+y}$$
From here, I leave the rest to you, as what follows is not difficult.
A: Let $u=\ln(1+1/x)$, so that $1+1/x=e^u$, or $x=1/(e^u-1)$, and the limit becomes
$$\lim_{u\to0^+}\left({1\over u}-{1\over e^u-1} \right)=\lim_{u\to0^+}{e^u-1-u\over u(e^u-1)}$$
One round of L'Hopital takes this to
$$\lim_{u\to0^+}{e^u-1\over(e^u-1)+ue^u}$$
and a second round to
$$\lim_{u\to0^+}{e^u\over e^u+e^u+ue^u}={1\over1+1+0}={1\over2}$$
