For $c > 0,$ find the limit, lim$_{n \to \infty} n(\sqrt[n]{c} - 1)$ 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. 
 A: Making the substitution of $x = \frac{1}{n}$, we can write,
$$\lim_{n\rightarrow\infty} n(c^{1/n} - 1) = \lim_{x\rightarrow 0^+} \frac{c^x - 1}{x} = \left.\frac{\mathrm{d}}{\mathrm{d}x} c^x\right|_{x=0} = \log(c)$$
A: 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,
$$\lim_{n \to \infty} n(\sqrt[n]{c}-1) = \frac{c-1}{c-1}\log c = \log c$$
A: 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:
$$\lim\limits_{n\rightarrow \infty}\frac{\frac{-\sqrt[n]{c}\log(c)}{n^2}}{\frac{-1}{n^2}}=\lim\limits_{n\rightarrow \infty}(\sqrt[n]{c}\log(c))=\log(c)$$
A: 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 
$$
\lim_{n\rightarrow\infty} a_n = \log c
$$
