Question

como se resolve:
lim (2^h-1)/h
h->0

Answer 1

Esse limite dá uma indeterminação da forma 0/0, portanto podemos utilizar L´Hôpital, derivando em cima e em baixo:

$\lim_{h \to 0} \frac{2^{h}-1}{h}= \lim_{h \to 0} \frac{2^{h}.ln2}{1}=2^{0}.ln2=ln2$

Answer 2

$\displaystyle\sf\lim_{ h \to 0}\dfrac{2^h-1}{h}\\\underline{\rm fac_{\!\!,}a}\\\sf t=2^h-1\longrightarrow h=\dfrac{\ell n(t+1)}{\ell n2}\\\sf h\to 0~se~ t \to 0\\\displaystyle\sf\lim_{ h \to 0}\dfrac{2^h-1}{h}=\lim_{ t \to 0}\dfrac{t}{\frac{\ell n(t+1)}{\ell n2}}\\\displaystyle\sf\ell n(2)\lim_{t \to 0}\dfrac{t}{\ell n(t+1)}\\\displaystyle\ell n(2)\lim_{ t \to 0}\dfrac{\frac{1}{\diagup\!\!\!t}\cdot\diagup\!\!\!t}{\frac{1}{t}\ell n(t+1)}\\\displaystyle\ell n(2)\lim_{t \to 0}\dfrac{1}{\ell n(t+1)^{\frac{1}{t}}}\\\underline{\rm fac_{\!\!,}a}\\\sf \dfrac{1}{u}=t\longrightarrow u=\dfrac{1}{t}\\\sf t\to 0~quando~ u\to\infty\\\displaystyle\ell n(2)\lim_{u \to \infty}\dfrac{1}{\ell n\bigg(1+\frac{1}{u}\bigg)^u}\\\sf\ell n(2)\cdot\dfrac{\displaystyle \sf\lim_{ u\to\infty}1}{\displaystyle\sf\lim_{u \to \infty}\ell n\bigg(1+\frac{1}{u}\bigg)^u}\\\sf\ell n(2)\cdot\dfrac{1}{\ell n\bigg[\displaystyle\sf\lim_{u \to \infty}\bigg(1+\frac{1}{u}\bigg)^u\bigg]}=\ell n(2)\cdot\dfrac{1}{\ell n[e] }=\ell n(2)$