Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule?

Question

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?

Answer 1

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 $(4^x)'$.

Answer 2

You could try this. As $n \to \infty$, $$ 4^{1/n} = \exp\left(\frac{\log 4}{n}\right) = 1 + \frac{\log 4}{n} + O(1/n^2) \\ 4^{1/n}-1 = \frac{\log 4}{n} + O(1/n^2) \\ n\left(4^{1/n}-1\right) = \log 4 + O(1/n) \\ \lim_{n\to\infty} n\left(4^{1/n}-1\right) = \log 4 $$

Answer 3

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 $\ln(4)$.

Answer 4

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$.

$$ \begin{split} \lim_{n \to +\infty} \frac{4^{\frac{1}{n}}-1}{\frac{1}{n}}&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{1}{n}}\\ &=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\ &=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \lim_{n \to +\infty}\frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\ &= 1 \cdot \ln 4 = \ln 4 \end{split} $$

Answer 5

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 $.

Answer 6

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 $$ \lim\limits_{z\rightarrow\infty} z\left(\sqrt[z]4-1\right) = \frac{2}{1/\ln 2} = 2\ln 2 = \ln 4 $$