Question

lim x-> -1 (tg^3)(x+1/4)/(x+1)^3

Answer 1

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


Façamos a seguinte mudança de variável:

$u=\dfrac{x+1}{4}\;\;\Rightarrow\;\;x+1=4u\;\;\Rightarrow\;\;(x+1)^{3}=64u^{3}$


e temos que $u\to 0$ quando $x\to -1.$ Substituindo, o limite fica


$L=\underset{u\to 0}{\mathrm{\ell im}}\;\dfrac{\mathrm{tg^{3}}(u)}{64u^{3}}\\ \\ \\ =\underset{u\to 0}{\mathrm{\ell im}}\;\dfrac{1}{64}\cdot \dfrac{\mathrm{sen^{3}}(u)}{u^{3}}\cdot \dfrac{1}{\cos^{3}(u)}\\ \\ \\ =\underset{u\to 0}{\mathrm{\ell im}}\;\dfrac{1}{64}\cdot \left[\dfrac{\mathrm{sen}(u)}{u} \right ]^{3}\cdot \dfrac{1}{\cos^{3}(u)}$


Agora que não há mais nenhuma indeterminação, chegamos a

$L=\dfrac{1}{64}\cdot \underset{u\to 0}{\mathrm{\ell im}}\;\left[\dfrac{\mathrm{sen}(u)}{u} \right ]^{3}\cdot \underset{u\to 0}{\mathrm{\ell im}}\;\dfrac{1}{\cos^{3}(u)}\\ \\ \\ =\dfrac{1}{64}\cdot \left[\underset{u\to 0}{\mathrm{\ell im}}\;\dfrac{\mathrm{sen}(u)}{u} \right ]^{3}\cdot \underset{u\to 0}{\mathrm{\ell im}}\;\dfrac{1}{\cos^{3}(u)}\\ \\ \\ =\dfrac{1}{64}\cdot 1^{3}\cdot \dfrac{1}{\cos^{3}(0)}\\ \\ \\ =\dfrac{1}{64}\cdot 1\cdot 1\\ \\ \\ =\dfrac{1}{64}\\ \\ \\ \\ \Rightarrow\;\;\boxed{\begin{array}{c} \underset{x\to -1}{\mathrm{\ell im}}\;\dfrac{\mathrm{tg^{3}}(\frac{x+1}{4})}{(x+1)^{3}}=\dfrac{1}{64} \end{array}}$