What is $\lim_{n\to\infty} n(2^{1/n} -1)$ How would you find the value of $\lim_{n\to\infty} n(2^{1/n} -1)$?
I tried taking the logarithm on both sides and it does not appear to be working.
 A: You can go this way.
$$ n(2^{1/n}-1 ) = n( e^{\frac{\ln(2)}{n}} - 1 ) = n \left( \left(1+\frac{\ln(2)}{n}+\frac{\ln^2(2)}{2!\,n^2}+\dots  \right)-1 \right)$$
$$ = {\ln(2)}+\frac{\ln^2(2)}{2!\,n}+\dots \longrightarrow_{n\to \infty} \ln(2) . $$
Note: You can use the big O notation if you are familiar with it

$$ e^{\frac{\ln(2)}{n}} = 1+\frac{\ln(2)}{n}+O\left( \frac{1}{n^2}\right)  $$

A: Write $f(x)=2^x=e^{x\log{2}}$ then $f'(x)=2^x\log{2}$. We have
$$n(2^{\frac{1}{n}}-1)=\frac{f(a_n)-1}{a_n}$$
Avec $a_n=\frac{1}{n}\rightarrow 0$. So 
$$\lim_{n\to \infty}n(2^{\frac{1}{n}}-1)=f'(0)=\log{2}$$
A: Set $ x = \dfrac{1}{n}$ , so $\lim_{n \to \infty} x = 0$
Therefore = 
$$\lim_{n\to \infty} n\left(2^{\frac{1}{n}}-1\right) = \lim_{x\to 0} \dfrac{\left(2^{x}-1\right)}{x} = \lim_{x\to 0} \ln 2 \dfrac{\left(e^{x\ln2}-1\right)}{x\ln2}$$
Now we do $m = x\ln2$ so $lim_{x \to 0} m = 0 \therefore$
$$\lim_{x\to 0} \ln 2 \dfrac{\left(e^{x\ln2}-1\right)}{x\ln2} = \ln 2\lim_{m\to 0} \dfrac{\left(e^{m}-1\right)}{m} $$
The limit is a classic and it's value is 1 (http://www.ies-math.com/math/java/calc/lim_e/lim_e.html), so 
$$\lim_{n\to \infty} n\left(2^{\frac{1}{n}}-1\right) = \ln 2$$
A: Let $x>1$. From $$\log(x)=\int_1^x\frac{dt}{t}$$ it follows that $$1-\frac{1}{x}<\log(x)<x-1$$ and $$\log(x)=n\log(x^{1/n}).$$ Then $$n(1-x^{-1/n})<\log(x)<n(x^{1/n}-1)$$ and so $$x^{-1/n} < \frac{\log(x)}{n(x^{1/n}-1)} < 1.$$ Now let $n\to\infty$.
