Question

lim x tends to 1 (x^3+3x^2-6x+2/x^3+3x^2-3x-1​

Answer 1

Answer:

$\binom{lim}{x \: tends \: to \: 1} \frac{ {x}^{3} + 3 {x}^{2} - 6x + 2 }{ {x}^{3} + 3 {x}^{2} - 3x - 1} \\ = \binom{lim}{x \: tends \: to \: 1} \frac{3 {x}^{2} + 6x - 6 }{3 {x}^{2} + 6x - 3 } \\ on \: putting \: x = 1 \\ = \frac{3 + 6 - 6}{3 + 6 - 3} \\ = \frac{3}{6} \\ = \frac{1}{2}$

Answer 2

Answer:

Step-by-step explanation: