Solve limit when $x\to \infty$ of $\ln$ expression

Question

I have this limit: $\lim \limits_{x \to \infty}x[\ln(x+1)-\ln(x)]$

Now I have tried to transform the expression to something like this: $\lim \limits_{x \to \infty}[\ln(\frac{x+1}{x})^x]$

I was thinking of making this look like the limit of number e, but doesn't look like it, I'm stuck. I would really appreciate the help!

Answer 1

You are on the right track. Note that, as $x\to +\infty$, $$\left(\frac{x+1}{x}\right)^x=\left(1+\frac{1}{x}\right)^x\to e$$ (as a reference see for example HERE).

Answer 2

Since $\ln((x+1)/x)=\ln(1+1/x)\rightarrow 0$, just write this as $\frac{\ln(x+1)-\ln(x)}{1/x}$ and apply l'Hopital's rule.

Answer 3

I though it might be instructive to provided a "HINT" so that evaluation of the limit relies on pre-calculus tools only.

HINT:

Use the inequalities

$$\frac1{x+1}\le \log\left(1+\frac1x\right)\le \frac1x$$

(See THIS ANSWER), multiply by $x$, and apply the squeeze theorem.