Limits, n-th root

Question

I have troubles with this limit: $\displaystyle \lim_{n\rightarrow \infty} n(\sqrt[n]{2}-1)$. It's probably about some trick, but I can't find it :-).

Thank you for your help

Answer 1

First method: Derivative of $2^x$: $$\lim_{n\to\infty}n(2^{1/n}-1)=\lim_{t\to 0}\frac{2^{t}-1}{t}.$$ With $f(t)=2^t$ and derivative at $0$ $\displaystyle\lim_{t\to0} \frac{2^{0+t}-2^0}{t}=\frac{2^t-1}{t} $. So all that is left is to compute $f'(0)$

Second method: L'Hopital: $$\lim_{n\to\infty}n(2^{1/n}-1)=\lim_{t\to 0}\frac{2^{t}-1}{t}\equiv \lim_{t\to 0} \log 2\times 2^t=\log 2$$ Third method: Taylor series: $$\lim_{n\to\infty}n(2^{1/n}-1)=\lim_{t\to 0}\frac{2^{t}-1}{t}=\lim_{t\to 0}\frac{1+t\log 2+O(t^2)-1}{t}=\lim_{t\to 0}\frac{t\log 2+O(t^2)}{t}=\log 2$$

Answer 2

Substitute $x=1/n$ to get $$\lim_{x\to 0}\frac{2^x-1}{x}$$ Recognize that this is, by definition, the derivative of $2^x$ evaluated at $x=0$.

Answer 3

Put $a = \sqrt[n]{2} \to \ln a = \dfrac{\ln 2}{n} \to n\left(\sqrt[n]{2}-1\right) = \dfrac{\ln 2\left(a-1\right)}{\ln a}$.

Claim: $\displaystyle \lim_{a\to 1^{+}} \dfrac{a-1}{\ln a} = 1$.

Proof: $f(x) = e^{x}$ on $(0,\infty)$ is countinuosly differentiable, so by MVT, $f(\ln a) - f(\ln 1) = f'(\alpha)\left(\ln a - \ln 1\right)$ for some $\alpha \in (\ln 1, \ln a) = (0, \ln a) \Rightarrow \dfrac{a-1}{\ln a} = e^{\alpha} \to e^0 = 1$ when $a \to 1^{+} \Rightarrow n\left(\sqrt[n]{2}-1\right) \to \ln 2$.