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.

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• 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 – Aleksandar Jul 17 '15 at 2:29
• Is it possible to do this using rules of sequences (i.e. not L'Hopital)? – user217285 Jul 17 '15 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/… – Paramanand Singh Jul 17 '15 at 5:07
• Perhaps this link might prove itself helpful... – Lucian Jul 17 '15 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. – hardmath Jul 17 '15 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? – Jellyfish Jul 17 '15 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. – Terra Hyde Jul 17 '15 at 2:48
• @Terra is using the rule $(a^x)' = \ln{a} \cdot a^x$ – user217285 Jul 17 '15 at 2:48
• @TerraHyde Thanks. Now I understand. – Jellyfish Jul 17 '15 at 15:23
• @Nitin Thanks for mentioning that formula. I seemed to forgot that somehow. – Jellyfish Jul 17 '15 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 $$\lim_{n\rightarrow\infty} a_n = \log c$$