Calculating $\lim_{h \to 0}\frac{1}{h}\ln\left(\frac{e^h-1}{h}\right)$ in an elementary way

Question

I have to calculate the limit of $$\lim_{h \to 0}\dfrac{1}{h}\ln\left(\dfrac{e^h-1}{h}\right)$$ I can calculate this limit using Taylor series, and got the answer $\frac{1}{2}$. However, I want to solve this limit in a somewhat "elementary" way, not using Taylor series, Laurent series, or L'Hopital's rule. Are there any such ways?

Answer 1

$$\lim_{h\to0}\frac1h\ln\left(\frac{e^h-1}h\right)=\left.\frac{d}{dx}\ln\left(\frac{e^x-1}x\right)\right|_{x=0}$$

Answer 2

An elementary method is to bound the limit from above and from below. Try and bound $\log{(e^h -1)\over h}$ between $\log(1+{h\over2})< \log{(e^h -1)\over h} < \log(1+{h\over2}+\alpha {h^2})$ when $h$ is small enough and some $\alpha$.

Answer 3

\begin{equation*} f(x)= \begin{cases} \frac{e^x-1}x&x\ne0\\ 1&x=0 \end{cases} \end{equation*}
\begin{align} \lim_{h \to 0}\dfrac{1}{h}\ln\left(\dfrac{e^h-1}{h}\right)&=\lim_{h \to 0}\dfrac{\ln f(h)-\ln f(0)}{h}\\ &=\left(\ln f(x)\right)'\left.\right|_{x=0}=\frac{f'(0)}{f(0)}=f'(0)\\ &=\lim_{x \to 0}\dfrac{\frac{e^x-1}x-1}{x}=\lim_{x \to 0}\dfrac{e^x-1-x}{x^2}=L\\ (x\rightarrow2x)\quad&=\lim_{x \to 0}\dfrac{e^{2x}-1-2x}{4x^2}\\ &=\lim_{x \to 0}\left(\dfrac{e^{2x}-e^x-x}{4x^2}+\dfrac{e^x-1-x}{4x^2}\right)\\ &=\lim_{x \to 0}\dfrac{e^{2x}-e^x-x}{4x^2}+\frac L4\\\\ \therefore L&=\lim_{x \to 0}\dfrac{e^{2x}-e^x-x}{3x^2}\\ (x\rightarrow2x)\quad&=\lim_{x \to 0}\dfrac{e^{4x}-e^{2x}-2x}{12x^2}\\ &=\lim_{x \to 0}\left(\dfrac{e^{4x}-3e^{2x}+2e^x}{12x^2}+\dfrac{e^{2x}-e^x-x}{6x^2}\right)\\ &=\lim_{x \to 0}\dfrac{e^x(e^x-1)^2(e^x+2)}{12x^2}+\frac L2\\ &=\lim_{x \to 0}\frac14\left(\frac{e^x-1}x\right)^2+\frac L2\\ &=\frac14\cdot\left(\left.\left(e^x\right)'\right|_{x=0}\right)^2+\frac L2\\\\ \therefore L&=\frac12\left(\left.\left(e^x\right)'\right|_{x=0}\right)^2=\frac12 \end{align}