Respostas
Resposta:
Calcular o limite
\Large\begin{array}{l} \lim\limits_{x\to 27}\dfrac{\sqrt[3]{x}-3}{x-27} \end{array}x→27limx−273x−3
Faça uma mudança de variável:
\Large\begin{array}{l} \sqrt[3]{x}=u\quad\Rightarrow\quad x=u^3 \end{array}3x=u⇒x=u3
e temos que u tende a 3 quando x tende a 27. Assim, o limite fica
\begin{lgathered}\Large\begin{array}{l} \lim\limits_{u\to 3}~\dfrac{u-3}{u^3-27}\\\\ \lim\limits_{u\to 3}~\dfrac{u-3}{u^3-3^3} \end{array}\end{lgathered}u→3lim u3−27u−3u→3lim u3−33u−3
Fatore a diferença entre dois cubos usando produtos notáveis:
a³ − b³ = (a − b) . (a² + ab + b²)
para a = u e b = 3:
\begin{lgathered}\Large\begin{array}{l} =\lim\limits_{u\to 3}~\dfrac{u-3}{(u-3)\cdot (u^2+u\cdot 3+3^2)}\\\\ =\lim\limits_{u\to 3}~\dfrac{u-3}{(u-3)\cdot (u^2+3u+9)} \end{array}\end{lgathered}=u→3lim (u−3)⋅(u2+u⋅3+32)u−3=u→3lim (u−3)⋅(u2+3u+9)u−3
Simplifique o fator comum (u − 3) que aparece no numerador e no denominador:
\begin{lgathered}\Large\begin{array}{l} =\lim\limits_{u\to 3}~\dfrac{1}{u^2+3u+9}\\\\ =\dfrac{1}{3^2+3\cdot 3+9}\\\\ =\dfrac{1}{9+9+9} \end{array}\end{lgathered}=u→3lim u2+3u+91=32+3⋅3+91=9+9+91
\Large\begin{array}{l} =\dfrac{1}{27} \end{array}=271 <———— esta é a resposta.
Bons estudos! :-)