<s>[INST] Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule? I can evaluate the limit with L'Hospital's rule:
$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=\ln4$
But is there any way to do it without using L'Hospital's rule?
 [/INST] Here's a trick to prove convergence of the continuous limit$$
\lim\limits_{x\rightarrow\infty} x \left(\sqrt[x]4 - 1\right)
$$
if you know also know how to integrate $2^x$.
Observe:\begin{eqnarray}
x \left(\sqrt[x]4 - 1\right) = x \left(\sqrt[x]2 - 1\right)\left(\sqrt[x]2 + 1\right) = 2x\left(\sqrt[2x]4 - 1\right)\frac{\left(\sqrt[x]2 + 1\right)}{2}
\end{eqnarray}
Thus we get $$
x\left(\sqrt[x]4 - 1\right) = \frac{\sqrt[x]2 + 1}2 2x\left(\sqrt[2x]4-1\right) \frac{\sqrt[x]2 + 1}2 \frac{\sqrt[2x]2 + 1}{2} 4x\left(\sqrt[4x]4-1\right) = \left(\prod_{k=0}^{n-1} \frac{\sqrt[2^kx]2+1}{2}\right) 2^n x \left(\sqrt[2^n x]4 -1\right)
$$
Note that the multiplicands are always bigger than $1$, so this implies that $2^n x \left(\sqrt[2^nx]4 - 1\right)$ is decreasing in $n$, so the limit exists for each $x$. It's also pretty clear that the function $x\to x\left(\sqrt[x]4-1\right)$ doesn't have any oscillations in the limit, hence we have that the continuous limit converges, so we just have to compute it for one value of $x$, say $x=1$. Thus we have $$
\lim\limits_{z\rightarrow\infty} z \left(\sqrt[z]4-1\right) = \lim\limits_{n\rightarrow\infty} 2^n \left(\sqrt[2^n]4-1\right) = \lim\limits_{n\rightarrow\infty} \frac{\sqrt[1]4-1}{\prod\limits_{k=0}^{n} \frac{\sqrt[2^k]2+1}{2}} = \frac2 {\prod\limits_{k=1}^\infty \frac{2^{1/2^k} + 1}{2}}
$$
Expanding the partial products, we can see that this product actually isn't hard to evaluate:\begin{eqnarray}
\prod_{k=1}^1 \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac12 + 1}2\\
\prod_{k=1}^2 \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac14 + 2^\frac12 + 2^\frac34 + 1}4\\
&\vdots&\\
\prod_{k=1}^N \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac1{2^N} + 2^\frac{2}{2^N} + \cdots + 2^\frac{2^N-1}{2^N} + 2^\frac{2^N}{2^N}}{2^N}
\end{eqnarray}
The RHS is a left Riemann sum for $\int_0^1 2^t dt$ splitting the interval into $2^N$ subintervals. Thus we get $$
\prod_{k=1}^\infty \frac{2^{\frac1{2^k}} + 1}{2} = \int_0^1 2^t dt = \frac{2-1}{\ln 2} = \frac1{\ln 2}
$$
Hence we conclude $$
\lim\limits_{z\rightarrow\infty} z\left(\sqrt[z]4-1\right) = \frac{2}{1/\ln 2} = 2\ln 2 = \ln 4
$$
</s>