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?
 A: If $f(x)=4^x$, then $f'(x)=\log(4)4^x$ and, in particular, $f'(0)=\log(4)$. In other words,$$\lim_{h\to0}\frac{4^h-1}h=\log(4)$$and therefore$$\lim_{n\to\infty}\frac{4^{1/n}-1}{1/n}=\log(4),$$which is the same thing as asserting that$$\lim_{n\to\infty}n\left(\sqrt[n]4-1\right)=\log(4).$$Note that all that I used was the definition of derivative together with the knowledge of $(4^x)'$.
A: You could try this.  As $n \to \infty$,
$$
4^{1/n} = \exp\left(\frac{\log 4}{n}\right)
= 1 + \frac{\log 4}{n} + O(1/n^2)
\\
4^{1/n}-1 = \frac{\log 4}{n} + O(1/n^2)
\\
n\left(4^{1/n}-1\right) = \log 4 + O(1/n)
\\
\lim_{n\to\infty} n\left(4^{1/n}-1\right) = \log 4
$$
A: You could use a Maclaurin expansion for $e^x$:
$$\begin{align}
n\left(\sqrt[n]{4}-1\right)
&=n\left(e^{\ln(4)/n}-1\right)\\
&=n\left(\frac{\ln(4)}{n}+\frac{1}{2}\left(\frac{\ln(4)}{n}\right)^2+\cdots\right)\\
&=\ln(4)+\frac{1}{2}\frac{\ln(4)^2}{n}+\frac{1}{6}\frac{\ln(4)^3}{n^2}+\cdots\\
\end{align}$$
As $n\to\infty$, convince yourself that the trailing terms collectively converge to $0$, leaving $\ln(4)$.
A: Another approach using known limit

$$ \lim_{n \to +\infty} \frac{\mathrm{e}^{a_n}-1}{a_n} = 1 $$
where $a_n$ is a sequence such that  $ \lim_{n \to +\infty} a_n = 0$.

$$
\begin{split}
\lim_{n \to +\infty} \frac{4^{\frac{1}{n}}-1}{\frac{1}{n}}&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{1}{n}}\\
&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\
&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \lim_{n \to +\infty}\frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\
&=  1 \cdot \ln 4 = \ln 4
\end{split}
$$
A: Let $t=\frac 1n \to 0$: $$\lim_{t\to 0} \frac{4^t-1}{t}  $$ which is of the well-known form $\lim_{x\to 0} \frac{a^x-1}{x} =\ln a $.
A: 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 $$
\lim\limits_{z\rightarrow\infty} z\left(\sqrt[z]4-1\right) = \frac{2}{1/\ln 2} = 2\ln 2 = \ln 4
$$
