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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!

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Just remember to put in $\LaTeX$ in between dollar signs on this website. – mathguy Sep 13 '12 at 22:24
Maple says ln(2) – i. m. soloveichik Sep 13 '12 at 22:25
Youu can rewrite your expression $$\frac{2^{1/n}-1}{1/n}$$. Then it comes easily. – S4M Sep 13 '12 at 22:46

It should be noted that we can define the (natural) logartihm 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.

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Hint $$\lim _ {x\rightarrow 0} \frac{2^x-2^0}{x-0}=\ln2$$

ADDED:$$2^{\frac{1}{n}}=e^{\frac{\ln2}{n}}=1+\frac{\ln2}{n}+\mathrm{O}\left ( \frac{1}{n^2} \right )$$

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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 $f'(x)=\ln 2 2^x$, then you have $$\lim\limits_{n\to\infty} \frac{2^{1/n}-1}{1/n}= f'(0)=\ln 2.$$

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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

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Or just use the definition of the derivative for $2^x$. – Erick Wong Sep 13 '12 at 22:51

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

$$\lim_{n\to\infty}f(n)\,\,,\,n\in\Bbb N\,$$

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