Finding $\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$ I am trying to find the limit of:
$$\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$$
I know it should be very simple but I don't seem to get it.
Thanks in advance for any help!
 A: It should be noted that we can define the (natural) logarithm of a number $x>0$ as
$$L(x)=\lim_{h\to 0}\frac{x^h-1}{h}$$
Although it may take a time, it is not awfully complicated to prove this function indeed has the defining properties of the logarithm.
$$\tag 1 L(xy)=L(x)+L(y)$$
$$\tag 2 L(x^a)=a L(x)$$
$$\tag 3 1-\frac 1 x \leq L(x)\leq x-1$$
$$\tag 4\lim_{x\to 0}\frac{L(x+1)}{x}=1$$
$$\tag 5 L'(x)=\frac 1 x$$
Note that $1\Rightarrow 2 \;(a\in \Bbb Z)\Rightarrow 3\Rightarrow 4\Rightarrow 5$.
On the other hand, a naïve use of L'Höpital's rule, gives
$$\lim_{h\to 0}\frac{x^h-1}{h}\mathop=\limits^{\frac 0 0}\lim_{h\to 0}\frac{x^h\log x}{1}=\log x$$
If you're interested in the proofs, just let me know.
A: There are several answers and comments suggesting L'Hospital's
rule.
In fact, it is a bit easier - you only need the definition of the
derivative.
$$\lim\limits_{n\to\infty} \frac{2^{1/n}-1}{1/n}= \lim\limits_{x\to0} \frac{2^x-2^0}{x-0}$$
So this expression is precisely the value of the derivative of
$f(x)=2^x$ at $x=0$.
If you already know that the derivative of $f(x)=2^x$ is
$f'(x)= 2^x  \ln 2$, then you have
$$\lim\limits_{n\to\infty} \frac{2^{1/n}-1}{1/n}= f'(0)=\ln 2.$$
A: Hint $$\lim _ {x\rightarrow 0} \frac{2^x-2^0}{x-0}=\ln2$$
ADDED:$$2^{\frac{1}{n}}=e^{\frac{\ln2}{n}}=1+\frac{\ln2}{n}+\mathrm{O}\left ( \frac{1}{n^2} \right )$$
A: Hint: Write the limit as $\lim_{n\to\infty}\frac{\sqrt[n]{2}-1}{\frac{1}{n}}$
and note that $\lim_{n\to\infty}\sqrt[n]{2}=1$. Now use L'Hospital's
Rule and clark's hint.
A: Take the function $\,f(x):=x(\sqrt[x] 2-1)\,$ and let us apply L'Hospital:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{2^{1/x}-1}{1/x}\stackrel{\text{L'H}}=\lim_{x\to\infty}\frac{\left(-\frac{1}{x^2}\right)\log 2\cdot 2^{1/x}}{\left(-\frac{1}{x^2}\right)}=\lim_{x\to\infty}\log 2\cdot 2^{1/x}=\log 2$$
Of course, the limit will be the same if we take 
$$\lim_{n\to\infty}f(n)\,\,,\,n\in\Bbb N\,$$
