Limit without using Heine or L'Hôpital

Question

How to find the $\lim_{n\to\infty}n(\sqrt[n]{e}-1)$ ? I can't use Heine theorem neither L'Hôpital's rule.

Answer 1

$$\sqrt[n]e = e^{\frac1n} = 1 + \frac1n + o\left(\frac1{n}\right)$$

So $n (\sqrt[n] e-1) = 1 + o(1) \to 1$.

Answer 2

Hint:

$$ \frac{e^{\frac{1}{n}}-1}{\frac{1}{n}} = \frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}-0} $$ for $f$ defined as $f(x)=e^x$. Does the above form ring a bell? Something like $\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ should.

Answer 3

More generally, to get $\lim_{n\to\infty}n(\sqrt[n]{a}-1) $ where $a > 0$, this is

$\begin{array}\\ \lim_{n\to\infty}n(\sqrt[n]{a}-1) &=\lim_{n\to\infty}n(e^{\ln a/n}-1)\\ &=\lim_{n\to\infty}n(1+\ln a/n+O(1/n^2)-1)\\ &=\lim_{n\to\infty}\ln a+O(1/n)\\ &=\ln a\\ \end{array} $

Answer 4

Rewrite the expression as $\;\dfrac{\mathrm e^{\tfrac1n}-\mathrm e^0}{\dfrac1n}$, a rate of variation of the exponential function from $0$. hence the limit is the derivative of $\mathrm e^x$ at $x=0$, i.e. $1$.