Question

Evaluate limx->-1 x(cube) +1/ x + 1

Answer 1

Hey there!

$\lim_{x \to -1 } \frac { x^3 + 1 } {x+ 1 }$

= $\lim_{x \to -1 } \frac { (x+1)(x^2 -x+1) } {x+1 }$
= $\lim_{x \to -1} [{x^2-x+1} ]$
= (-1)²-(-1)+1
= 1 +1 + 1
= 3 .

Hope helped!

Answer 2

\lim_{x \to -1 } \frac { x^3 + 1 } {x+ 1 }lim​x→−1​​​x+1​​x​3​​+1​​ 

= \lim_{x \to -1 } \frac { (x+1)(x^2 -x+1) } {x+1 }lim​x→−1​​​x+1​​(x+1)(x​2​​−x+1)​​ 
= \lim_{x \to -1} [{x^2-x+1} ]lim​x→−1​​[x​2​​−x+1] 
= (-1)²-(-1)+1
= 1 +1 + 1 
= 3 .
hope it will help you