Question

Calcular el siguiente límite indeterminado de la forma 0/0

$\lim_{x \to 4} \frac{\sqrt{x^{3}-7x }-6 }{x^{2}-4x }$

Answer 1

Respuesta:

$\frac{41}{48}$

Explicación paso a paso:

$\lim_{x \to 4} \frac{\sqrt{x^{3}-7x }-6 }{x^{2}-4x } \\ \\ \lim_{x \to 4} \frac{(\sqrt{x^{3}-7x }-6)(\sqrt{x^{3}-7x } + 6)}{( {x}^{2} - 4x)(\sqrt{x^{3}-7x } + 6)} \\ \\ \lim_{x \to 4} \frac{ {(\sqrt{x^{3}-7x })}^{2} - {(6)}^{2} }{x(x - 4)(\sqrt{x^{3}-7x } + 6)} \\ \\ \lim_{x \to 4} \frac{ {x}^{3} - 7x - 36 }{x(x - 4)(\sqrt{x^{3}-7x } + 6)} \\ \\ \lim_{x \to 4} \frac{(x - 4)( {x}^{2} + 4x + 9) }{x(x - 4)(\sqrt{x^{3}-7x } + 6)} \\ \\ \lim_{x \to 4} \frac{ {x}^{2} + 4x + 9}{x( \sqrt{ {x}^{3} - 7x } + 6)} \\ \\ \frac{\lim_{x \to 4} {x}^{2} + 4x + 9 }{\lim_{x \to 4} x(\lim_{x \to 4} \sqrt{ {x}^{3} - 7x} + 6)} \\ \\ \frac{ {4}^{2} + 4(4) + 9 }{4( \sqrt{ {4}^{3} - 7(4) } + 6) } \\ \\ \frac{16 + 16 + 9}{4 (\sqrt{64 - 28} + 6)} \\ \\ \frac{ 41}{(4)(6+ 6)} \\ \\ \frac{ 41}{(4)(12)} \\ \\ \frac{ 41}{48}$