How to solve $\lim_{{x \to +\infty}} x(x+1) \ln \left(\frac{x+1}{x}\right) - x=\frac12$ without using Taylor expansion

Question

I have made numerous attempts to prove that the limit $\lim_{{x \to +\infty}} x(x+1) \ln \left(\frac{x+1}{x}\right) - x$ equals $\frac{1}{2}$ without resorting to Taylor expansion. Unfortunately, all my efforts have ended in failure

Attempt:

Given the original limit:

$\lim_{{x \to +\infty}} x(x+1) \ln \left(\frac{x+1}{x}\right) - x$

Substituting $y = \frac{1}{x}$ gives us:

$\lim_{{y \to 0^+}} \frac{1}{y} \left(\frac{1}{y} + 1\right) \ln \left(\frac{1 + 1/y}{1/y}\right) - \frac{1}{y}$

Simplify this expression:

$\lim_{{y \to 0^+}} \frac{1 + y}{y^2} \ln(y+1) - \frac{1}{y}$

Using L'Hôpital's Rule for the indeterminate form $\infty \cdot 0 - \infty$:

$\lim_{{y \to 0^+}} \frac{\ln(y+1)}{1/y}$

Differentiating the numerator and denominator separately:

$\lim_{{y \to 0^+}} \frac{\frac{1}{y+1}}{-1/y^2}$

Simplify this expression:

$\lim_{{y \to 0^+}} -y = 0$

Therefore, after applying the variable change and using L'Hôpital's Rule, the limit evaluates to $0$.

Answer 1

$$\lim_{x\to +\infty}x(x+1)\log\left(1+1/x\right)-x=\lim_{y\to 0^+}\left(\frac{\log (1+y)}{y}+\frac{\log(1+y)-y}{y^2}\right)=1-1/2=1/2$$ Note that: $$\lim_{y\to 0}\frac{\log(1+y)}{y}=\lim_{x\to 0}\frac{x}{e^x-1}=1 $$ To solve the second limit without Taylor expansion (and without L'Hopital's rule), you can refer to this question.

Answer 2

I think that you can not use L'Hopital in the indetermination that you have used it. Try this:

$$ \lim_{y\rightarrow 0}\frac{(1+y)\ln(1+y)}{y^{2}}-\frac{1}{y}=\lim_{y\rightarrow 0}\frac{(1+y)\ln(1+y)-y}{y^{2}}=\frac{1}{2} $$ where the L'Hopital rule has been used in the second limit

Answer 3

You used L'Hôpital's Rule incorrectly - $\frac{\ln((y+1)}{y^{-1}}$ has the numerator going to 0 and the denominator going to $\infty$, so you can't use L'Hôpital's Rule.
Instead try to write the entire expression as 1 fraction, i.e. $$\frac{1+y}{y^2}\ln(y+1)-\frac{1}{y}=\frac{(1+y)\ln(y+1)-y}{y^2}$$ Notice that the numerator and the denominator go to 0 as $y$ approaches 0. Applying L'Hôpital's Rule to this expression twice (once you differentiated once you should notice that the new fraction also has both the numerator and denominator going to 0, so that you can apply L'Hôpital's Rule again), and you'll get that the limit is $\frac{1}{2}$