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How to compute the following limit?

$\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)$

We know that it's got something to do with $\ln$ or $\exp$.

We know that $\lim\limits_{n\to\infty} \sqrt[n]{a} = 1$ but it seems to not to be true that therefore $\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1) = \lim\limits_{n\to\infty} n·(1-1) = 0$.

What we know is

$\lim\limits_{n\to\infty} (1-\frac{1}{n})^n = \lim\limits_{n\to\infty} (1 + \frac{1}{n})^n \lim\limits_{n\to\infty} (1-\frac{1}{n})^{n+1} = e$


$\lim\limits_{n\to\infty} (1+\frac{x}{n})^n = \exp(x)$

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Hint: You can use the fact that $\lim\limits_{x\to0} \frac{a^x-1}x$ is precisely the derivative of the function $f(x)=a(x)$. (In case you've already learned about derivatives and are allowed to use them.) –  Martin Sleziak Mar 26 '12 at 14:27

4 Answers 4

up vote 4 down vote accepted

Substitute $t=\frac{1}{n}$ , hence :

$L=\displaystyle \lim_{t \to 0} \frac{a^t-1}{t}$

Now , apply L'Hopital rule ,hence :

$L=\displaystyle \lim_{t \to 0} a^t \cdot \ln a=\ln a$

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Actually $L$ is the definition of the derivative of $a^{t}$ at 0. You should get it from the definition, since applying L'H in this case would be circular logic.... –  N. S. Mar 26 '12 at 15:03
@N.S. This was one tricky thing I learned in my first year of university. Applying l'Hospital to basic limits is circular logic because you literally use the value of the limit when proving the formula for the derivative. –  Beni Bogosel Mar 26 '12 at 15:16

Here's how Wolfram|Alpha solves it.

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the edit broke the link. –  Tyler Mar 26 '12 at 15:10

Set $x_n=n(\sqrt[n]{a}-1)$ for all $n$. Rearranging this you get $(\frac{x_n}{n}+1)^n=a$ for all $n \in \mathbb{N}$. But this also means that $$(\frac{x_n}{n}+1)^n \rightarrow a=e^{\ln(a)}$$ as $n\rightarrow \infty$. Hence we must have $x_n \rightarrow \ln(a)$ as $n \rightarrow \infty$.

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Nice solution. The last part, namely that $x_n \to \ln(a)$ needs a little justification though.. It follows easely if you observe that $\ln(a)$ is the only limit point of $x_n$. –  N. S. Mar 26 '12 at 15:05
Thanks :) I left it out intentionally. I think it can be instructive for the OP the fill in the details. But never the less you are of course correct. Formally, if $x_n$ do not converge to $ln(a)$, then since the sequence of $x_n$:s obviously is bounded, we could find a subsequence converging to a number c not equal to ln(a), which is a contradiction –  user22705 Mar 26 '12 at 16:01

Using Taylor's expansion, $$ n(\sqrt[n]{a}-1)=n(e^{\frac1n\log a}-1)=n(1+\frac1n\,\log a+o(\frac1{n^2})-1)=\log a + o(\frac1n)\xrightarrow{n\to\infty}\log a. $$

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