Why does this tend to $\ln|2|$

Question

May I know why is this true?
When $k \to 0$, $\frac{2^k -1}{k} \to \ln|2| $.
From what I can see, this fraction would tend to $\frac{0}{0}$, which has no exact value. And if you use L'Hospital's law, it still does not tend to $\ln|2|$.
Thank you very much for you guys reply.

Answer 1

By L'Hospital's rule,

$$ \lim\limits_{k \to 0} \frac{2^k -1}{k} = \lim\limits_{k \to 0} \frac{(\ln 2) (2^k) }{1} = \ln 2. $$ Since the second limit converges, the first converge to the same value.

Based on your comments, it appears that you are applying the formula for the derivative of a power of some variable $x$, $$ \frac{d}{dx} x^n = nx^{n-1}\quad \text{ where $n\in \Bbb R$ } $$

when you in fact are dealing with an exponential of $x$, which satisfies $$ \frac{d}{dx} a^x = (\ln a)x^{a}\quad \text{ where $a$ > 0} $$

Answer 2

set $2^{k}-1=t$

then we have: $$\lim_{k\to0}\frac{2^{k}-1}{k}=\lim_{t\to0}\frac{t}{\log_{2}\left(t+1\right)}=\lim_{t\to0}\frac{1}{\frac{\log_{2}\left(t+1\right)}{t}}=\frac{1}{\log_{2}\left(\lim_{t\to0}(t+1\right))^{\large\frac{1}{t}}}=\frac{1}{\log_{2}\left(e\right)}=\ln\left(2\right)$$

Here I used the fact that $\log_{2}\left(u\right)$ is continuous over its domain.

Answer 3

This is actually the rate of variation from the value $x=0$ of the function $x\mapsto 2^x$. By definition, this is the derivative of $2^x$ at $x=0$. Now, for any $a>0$, $$(a^x)'=\bigl(\mathrm e^{x\ln a}\bigr)'=\mathrm e^{x\ln a}\ln a=a^x\ln a.$$