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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!

  • 6
    Youu can rewrite your expression $\frac{2^{1/n}-1}{1/n}$. Then it comes easily.2012-09-13

5 Answers 5

5

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.

  • 0
    @Did Yes, indeed. :)2016-06-09
4

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

3

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

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

1

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.

  • 0
    Invoking L'Hopital to prove that the limit of $(f(t)-f(0))/t$ when $t\to0$ is $f'(0)$ is, strictly speaking, making a logical circle.2016-06-09
1

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

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

  • 0
    Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/41874/discussion-on-answer-by-donantonio-finding-lim-n-to-infty-n-sqrtn2). Cool it, dudes.2016-06-30