# For $c > 0,$ find the limit, lim$_{n \to \infty} n(\sqrt[n]{c} - 1)$ [duplicate]

For $c > 0,$ find the limit, lim$_{n \to \infty} n(\sqrt[n]{c} - 1)$

Ok, I am not exactly on sure how to do this. Though here are some of my thoughts:

lim$_{n \to \infty} (nc^{1/n} - n) = nc^0 - n = 0,$ and I don't think this can be correct logically. Also, I think we can make it a $\log$ function by doing it like this: $(nc^{1/n} - n) = e^{\log(nc^{1/n} - n)}.$ But, then what? We can't distribute the log inside because that's not correct. I am lost. Any help?

Update: please don't mark this as a duplicate of this, since the other question was not required to use L'hopital's rule, but I need to use that in my question.

• However it appears as if the limit is $\log(c)$. wolframalpha.com/input/?i=lim+as+n-%3E+infinity+n%28c^%281%2Fn%29-1%29 Jul 17, 2015 at 2:29
• Is it possible to do this using rules of sequences (i.e. not L'Hopital)? Jul 17, 2015 at 2:49
• @Nitin: Using rules of sequences it is possible to prove that for $c > 0$ the limit exists and therefore defines a function of real variable $c$. If we call this function $f(c)$ then it can be further proved that $f(ab) = f(a) + f(b), f(a/b) = f(a) - f(b), f(1) = 0$ and that $f(x)$ is differentiable with $f'(x) = 1/x$. None of these proofs require any existing knowledge of $\log x$ and $e^{x}$. See paramanands.blogspot.com/2014/05/… Jul 17, 2015 at 5:07
• Perhaps this link might prove itself helpful... Jul 17, 2015 at 9:09
• @Jellyfish: In what way are you saying L'Hopital's Rule needs to be used in your Question? Your thoughts on the problem did not include any attempt to use L'Hopital's Rule. Jul 17, 2015 at 17:32

Making the substitution of $x = \frac{1}{n}$, we can write, $$\lim_{n\rightarrow\infty} n(c^{1/n} - 1) = \lim_{x\rightarrow 0^+} \frac{c^x - 1}{x} = \left.\frac{\mathrm{d}}{\mathrm{d}x} c^x\right|_{x=0} = \log(c)$$

Observe that $$c-1 = (\sqrt[n]{c}-1)\left(c^\frac{n-1}{n} + c^\frac{n-2}{n} + \cdots + 1\right)$$

so we can rewrite the limit as

$$\lim_{n\to\infty} n(\sqrt[n]{c}-1) = (c-1)\lim_{n\to\infty} \frac{n}{c^\frac{n-1}{n} + c^\frac{n-2}{n} + \cdots + 1}$$

But, inverting the right-hand side and interpreting it as a Riemann sum, we have

$$\lim_{n\to\infty} \frac{c^\frac{n-1}{n} + c^\frac{n-2}{n} + \cdots + 1}{n} = \int_0^1 c^x\;dx = \frac{c-1}{\log c}$$

so, altogether,

$$\lim_{n \to \infty} n(\sqrt[n]{c}-1) = \frac{c-1}{c-1}\log c = \log c$$

Notice first that this is better handled as a product: $$\lim\limits_{n\rightarrow \infty}(\sqrt[n]{c}-1)=0$$ so that: $$\lim\limits_{n\rightarrow \infty}n(\sqrt[n]{c}-1)=\infty\cdot0$$ By clever re-write, we have: $$\lim\limits_{n\rightarrow \infty}\frac{\sqrt[n]{c}-1}{\frac1n}=\frac00$$ And we can apply L'Hopital's rule to this: $$\lim\limits_{n\rightarrow \infty}\frac{\frac{-\sqrt[n]{c}\log(c)}{n^2}}{\frac{-1}{n^2}}=\lim\limits_{n\rightarrow \infty}(\sqrt[n]{c}\log(c))=\log(c)$$

• Sorry, if it's a stupid question, but how did you get $log(c )$ as a numerator? Jul 17, 2015 at 2:41
• We can write $\sqrt[n]{c}=c^{\frac1n}=e^{\frac{\log(c)}{n}}$ and differentiating requires the chain rule, which introduces that factor. Jul 17, 2015 at 2:48
• @Terra is using the rule $(a^x)' = \ln{a} \cdot a^x$ Jul 17, 2015 at 2:48
• @TerraHyde Thanks. Now I understand. Jul 17, 2015 at 15:23
• @Nitin Thanks for mentioning that formula. I seemed to forgot that somehow. Jul 17, 2015 at 19:04

i am not sure of the precise justification of the following argument, but it has the merit of motivating the result, and perhaps as a mnemonic:

LEMMA if $a_n$ is a convergent sequence \begin{align} \lim_{n\rightarrow\infty} a_n & = \log \lim_{n\rightarrow\infty} e^{a_n} \\ & = \log \lim_{n\rightarrow\infty} \left(1+\frac{a_n}{n} \right)^n \end{align}

with $a_n=n(\sqrt[n]{c}-1)$ this gives $$\lim_{n\rightarrow\infty} a_n = \log c$$