Question

lim x--> -3
(4^(x+3/5)-1)/(x+3)
resposta: 2/5 ln 2

Answer 1

$\underset{x \to -3}{\mathrm{\ell im}}\;\dfrac{4^{\frac{x+3}{5}}-1}{x+3}\\ \\ \\ =\underset{x \to -3}{\mathrm{\ell im}}\;\dfrac{(2^{2})^{\frac{x+3}{5}}-1}{x+3}\\ \\ \\ =\underset{x \to -3}{\mathrm{\ell im}}\;\dfrac{2^{\frac{2\,(x+3)}{5}}-1}{x+3}$


Fazendo a substituição

$\dfrac{2(x+3)}{5}=u\;\;\Rightarrow\;\;2(x+3)=5u\;\;\Rightarrow\;\;x+3=\dfrac{5u}{2}\\ \\ \\ x \to -3\;\;\Rightarrow\;\;u\to 0$

chegamos a

$\underset{x \to -3}{\mathrm{\ell im}}\;\dfrac{2^{\frac{2\,(x+3)}{5}}-1}{x+3}\\ \\ \\ =\underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{2^{u}-1}{\frac{5u}{2}}\\ \\ \\ =\dfrac{2}{5}\cdot \underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{2^{u}-1}{u}$


Fazendo outra substituição:

$v=2^{u}-1\;\;\Rightarrow\;\;2^{u}=1+v\;\;\Rightarrow\;\;(e^{\mathrm{\ell n\,}2})^{u}=1+v\\ \\ e^{u\mathrm{\,\ell n\,}2}=1+v\\ \\ u\mathrm{\,\ell n\,2}=\mathrm{\ell n\,}(1+v)\;\;\Rightarrow\;\;u=\dfrac{\mathrm{\ell n\,}(1+v)}{\mathrm{\ell n\,}2}\\ \\ \\ u \to 0\;\;\Rightarrow\;\;\;v\to 2^{0}-1\;\;\Rightarrow\;\;\;v\to 0$


temos

$\dfrac{2}{5}\cdot \underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{2^{u}-1}{u}\\ \\ \\ =\dfrac{2}{5}\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{v}{[\frac{\mathrm{\ell n\,}(1+v)}{\mathrm{\ell n\,}2}]}\\ \\ \\ =\dfrac{2}{5}\cdot \mathrm{\ell n\,}2\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{v}{\mathrm{\ell n\,}(1+v)}\\ \\ \\ =\dfrac{2}{5}\cdot \mathrm{\ell n\,}2\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{1}{\frac{1}{v}\mathrm{\,\ell n\,}(1+v)}\\ \\ \\ =\dfrac{2}{5}\cdot \mathrm{\ell n\,}2\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{1}{\mathrm{\,\ell n\,}(1+v)^{1/v}}$

$=\dfrac{2}{5}\cdot \mathrm{\ell n\,}2\cdot \dfrac{1}{\mathrm{\,\ell n\,}e}\\ \\ \\ =\dfrac{2}{5}\cdot \mathrm{\ell n\,}2\cdot \dfrac{1}{1}\\ \\ \\ =\dfrac{2}{5}\cdot \mathrm{\ell n\,}2$


$\boxed{\underset{x \to -3}{\mathrm{\ell im}}\;\dfrac{4^{\frac{x+3}{5}}-1}{x+3}=\dfrac{2}{5}\cdot \mathrm{\ell n\,}2}$