Question

lim x--> 2 (5^x-25)/(x-2)
resposta: 25ln5

Answer 1

$L=\underset{x \to 2}{\mathrm{\ell im}}\;\dfrac{5^{x}-25}{x-2}$


Substituição:

$x-2=u\;\;\Rightarrow\;\;x=u+2$

$u \to 0$ quando $x \to 2$


Substituindo, temos

$L=\underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{5^{u+2}-25}{u}\\ \\ \\ L=\underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{5^{u}\cdot 5^{2}-25}{u}\\ \\ \\ L=\underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{5^{u}\cdot 25-25}{u}\\ \\ \\ L=\underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{25\cdot (5^{u}-1)}{u}\\ \\ \\ L=25\cdot \underset{u \to 0}{\mathrm{\ell im}}\;\dfrac{5^{u}-1}{u}$


Substituição

$\mathbf{5^{u}-1=v}\;\;\Rightarrow\;\;5^{u}=v+1\;\;\Rightarrow\;\;(e^{\mathrm{\ell n\,}5})^{u}=v+1\\ \\ \Rightarrow\;\;e^{u\,\mathrm{\ell n\,}5}=v+1\;\;\Rightarrow\;\;u\mathrm{\,\ell n\,}5=\mathrm{\ell n\,}(v+1)\;\;\Rightarrow\;\;\mathbf{u=\dfrac{\mathbf{\ell n\,}(v+1)}{\mathbf{\ell n\,}5}}$

$v\to 0$ quando $u \to 0$


Então, temos

$L=25\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{v}{(\frac{\mathrm{\ell n\,}(v+1)}{\mathrm{\ell n\,}5})}\\ \\ \\ L=25\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{v\,\mathrm{\ell n\,}5}{\mathrm{\ell n\,}(v+1)}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{v}{\mathrm{\ell n\,}(v+1)}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{1}{\frac{1}{v}\cdot \mathrm{\ell n\,}(v+1)}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\cdot \underset{v \to 0}{\mathrm{\ell im}}\;\dfrac{1}{\mathrm{\ell n\,}(v+1)^{1/v}}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\cdot \dfrac{1}{\mathrm{\ell n\;}\underset{v \to 0}{\mathrm{\ell im}}\,(v+1)^{1/v}}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\cdot \dfrac{1}{\mathrm{\ell n\,}e}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\cdot \dfrac{1}{1}\\ \\ \\ L=25\,\mathrm{\ell n\,}5\\ \\ \\ \\ \Rightarrow\;\;\boxed{ \begin{array}{c} \underset{x \to 2}{\mathrm{\ell im}}\;\dfrac{5^{x}-25}{x-2}=25\,\mathrm{\ell n\,}5 \end{array} }$