Without using L'Hospital's rule and series expansion solve $e\lim\limits_{n \to \infty}\frac{e^{n\ln(1+1/n)-1}-1}{n\ln(1+1/n)-1}n(n\ln(1+1/n)-1)$

Question

This is my problem:

Solve only with the basic rules of natural logarithm and the definition for $e$.

$\lim\limits_{n\to\infty}(n(1+\frac{1}{n})^n - ne)$

And I got till here:

$e\lim\limits_{n \to \infty}\frac{e^{n\ln(1+\frac{1}{n})-1}-1}{n\ln(1+\frac{1}{n})-1}n(n\ln(1+\frac{1}{n})-1)$.

Both of these equations are equivalent with the limit $\frac{-e}{2}$. But I don't know how to transform to the next step. The first fraction should short itself giving me $1$ and the second multiplicand should give us $\frac{-1}{2}$ with the $e$ in front of the limit giving us the correct answer.

The definition for $e$ is $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.

But I think in this last step we should make a substitute $n = \frac{1}{x}$ and use the other definition $\lim\limits_{x\to0}(1+x)^{\frac{1}{x}} = e$

@vitamind Yes I know about that, but don't know how to make the connection I always get somewhere a $0$. — Account_For_Education, Jun 14 '21 at 14:10
This video by BlackPenRedPen provides a full solution to this problem : https://www.youtube.com/watch?v=FPHHv1UcrMA&t=602s — p_square, Jun 14 '21 at 14:40
@EdmondDantes Your title says: "Without using L'Hospital's rule". — vitamin d, Jun 14 '21 at 14:44
@EdmondDantes But without L'Hospital's rule, I don't think so this is possible — p_square, Jun 14 '21 at 14:53
Substituting $n=\frac1m$ transforms the limit to$$\lim_{m\to0}\frac{(1+m)^{\frac1m}-e}m$$which resembles the derivative of $f(m)=(1+m)^{\frac1m}$ at $m=0$. The derivative itself doesn't exist at this point, but $\lim\limits_{m\to0}f'(m)$ appears to have the same value of $-\frac e2$... — user170231, Jun 14 '21 at 15:46
@EdmondDantes You can probably prove this by finding good bounds on $\log$ and $\exp$. — vitamin d, Jun 14 '21 at 17:34

score 1 · Answer 1 · answered Jun 16 '21 at 01:51

Let me write found inequalities here $$\frac{1}{2n^2}- \frac{1}{n^3}\leqslant \frac{1}{n}-\ln\left(1+ \frac{1}{n}\right)\leqslant \frac{1}{2n^2}$$

p.s. I obtained expression identical to OP in previous question, but keep answer deleted, as hadn't inequalities in that time. We can prove them by examining the corresponding functions, but maybe someone will be able to see an elementary proof.

Without using L'Hospital's rule and series expansion solve $e\lim\limits_{n \to \infty}\frac{e^{n\ln(1+1/n)-1}-1}{n\ln(1+1/n)-1}n(n\ln(1+1/n)-1)$

1 Answers1