Question

Lim x4-16
x3-8
x--> 2

Answer 1

Hope it would be the correct answer

Answer 2

Hey there!

$\lim_{x\to2} \frac{ {x}^{4} - 16}{ {x}^{3} - 8} \\ \\ = \lim_{x\to2} \frac{( {x}^{2} + 4)( {x}^{2} - 4) }{(x - 2)( {x}^{2} + 2x + 4)} \\ \\ = \lim_{x\to2} \frac{( {x}^{2} + 4)( {x}- 2) (x + 2)}{(x - 2)( {x}^{2} + 2x + 4)} \\ \\ = \lim_{x\to2} \frac{( {x}^{2} + 4) \cancel{( x- 2)} (x + 2)}{ \cancel{(x - 2)}( {x}^{2} + 2x + 4)} \\ \\ = \lim_{x\to2} \frac{( {x}^{2} + 4) (x + 2)}{( {x}^{2} + 2x + 4)} \: \\ \\ = \lim_{x\to2} \frac{ {x}^{3} + 2 {x}^{2} + 4x + 8}{(x^{2} + 2x + 4)} \\ \\ = \frac{ {2}^{3 } + 2( {2}^{2} ) + 4(2) + 8 }{ {2}^{2} + 2(2) + 4 } \\ \\ = \frac{32}{12} \\ \\ = \frac{8}{3}$

Hope helped!