Limit $\lim \limits_{x\to0}{\frac{\ln(x+1)}{2^x-1}}$ wihout LHospital

Question

I want to find this limit but without using L'Hospital, with which is one-liner.

$$\lim \limits_{x\to0}{\frac{\ln(x+1)}{2^x-1}}$$

Answer 1

hint $$\ln{(1+x)}=x+o(x),x\to 0$$ because we know $$\dfrac{x}{x+1}<\ln{(1+x)}<x$$

$$2^x-1=\ln{2}\cdot x+o(x),x\to 0$$

Answer 2

As $x$ goes to $0$, we have $$\frac{\ln(x+1)}{e^{x \ln 2}-1}=\frac{x+O(x^2)}{x\ln 2+O(x^2)}\to \frac{1}{\ln 2}.$$

Answer 3

Consider $$\lim_{x\to 0} \frac{\ln(x+1)}{2^x - 1} = \lim_{x\to 0} \frac{\frac{1}{x}\ln(x+1)}{\frac{1}{x}(2^x - 1)}= \lim_{x\to 0} \frac{\frac{1}{x}\ln(x+1)}{\frac{1}{x}(2^x - 1)}.$$

Now, $$\lim_{x\to 0} \frac{1}{x}\ln(x+1) = \lim_{n\to\infty} \ln\left[(1+1/n)^n \right] = \ln(e) = 1$$ and $$\lim_{x\to 0} \frac{2^x -1}{x} = \left. \frac{d(2^x)}{dx}\right|_{x=0}=\ln(2)$$ and the result follows.

The second limit might also be one that you are expected to know independent of it being the derivative. In a sense, it follows directly from the other limit and it is certainly worth remembering.

Answer 4

Using only known limits: $$\begin{align}&\lim_{t \to 0}\frac{\ln(1 + t)}t = 1\\ &\lim_{t \to 0}\frac{e^t - 1}t = 1 \end{align}$$

and observing that $e^{x\ln 2} = 2^x$, rewrite as follows: $$\lim_{x \to 0}\frac{\ln(1 + x)}{2^x - 1} = \lim_{x \to 0}\frac{\ln(1 + x)}{x\ln 2}\cdot \frac{x\ln 2}{e^{x\ln 2} - 1} = \frac1{\ln 2}\cdot\lim_{x \to 0}\frac{\ln(1 + x)}{x}\cdot\lim_{x \to 0}\frac{x\ln 2}{e^{x\ln2} - 1} = \frac1{\ln 2}$$