Question

Вычислите:
$\lim_{x \to \ 64} \frac{ \sqrt{x} -8}{4- \sqrt[3]{x} }$

Answer 1

$\lim_{x \to 64} \frac{\sqrt{x}-8}{4-\sqrt[3]{x}}= \left[\begin{array}{ccc}\sqrt[6]{x}=t\\\sqrt[3]{x}=t^2\\\sqrt{x}=t^3\\x \to 64 =>t\to 2\end{array}\right] = \lim_{t \to 2} \frac{t^3-8}{4-t^2}= \lim_{t \to 2} -\frac{(t-2)(t^2+2t+4)}{t^2-4}$

$=\lim_{t \to 2} -\frac{(t-2)(t^2+2t+4)}{(t-2)(t+2)}=\lim_{t \to 2}-\frac{t^2+2t+4}{t+2}=-\frac{2*2+2*2+4}{2+2}=-\frac{12}4=-3$