Evaluating natural limit $\lim_{n \to \infty} \left( e^{2n} - 1\right) ^\frac{1}{n}$

Question

Any idea evaluating this $$ \lim_{n \to \infty} \left( e^{2n} - 1\right) ^\frac{1}{n} $$

after I raise all to e like so $$ \exp\left( \frac{\ln\left(e^{2n}-1\right)}{n}\right) $$ and Hopital's it I get stuck.

Thanks

Answer 1

You have $\displaystyle\lim_{n\to\infty}(e^{2n}-1)^{1/n}=\lim_{n\to\infty}\left(e^{2n}\bigg(1-\frac{1}{e^{2n}}\bigg)\right)^{1/n}=\lim_{n\to\infty}e^{2}\left(1-\frac{1}{e^{2n}}\right)^{1/n}=e^2(1^0)=e^2$

Answer 2

$$\dfrac{\ln(e^{2n}-1)}n=\dfrac{\ln(e^{2n})+\ln[1-e^{-2n}]}n=\dfrac{2n+\ln[1-e^{-2n}]}n$$

Now $\lim_{n\to\infty}e^{-2n}=\left[\lim_{n\to\infty}e^{-n}\right]^2=0^2$ as $e>1$

Answer 3

Much easier to use that $e^{2n}/2<e^{2n}-1<e^{2n}$ for $n>1$, and apply the squeeze theorem.

Answer 4

You can use L'Hospital's Rule directly as planned. To that end

$$\begin{align} \lim_{n\to \infty}\frac{\log(e^{2n}-1)}{n}&=\lim_{n\to \infty}\frac{2e^{2n}}{e^{2n}-1}\\\\ &=\lim_{n\to \infty}\frac{2}{1-e^{-2n}}\\\\ &=2 \end{align}$$

And we use the continuity of the exponential function to show that

$$\lim_{n\to \infty} (e^{2n}-1)^{1/n}=e^{\lim_{n\to \infty} \log(e^{2n}-1)^{1/n}}=e^2$$