Question

How do you find the limit of (2+x)^3 -8/x as the limit approaches 0? Please explain how you did it.​

Answer 1

Answer:  12

Step-by-step explanation:

Expand (2 + x)³, factor out the x, then replace x with zero.

$(x+2)^3 = (x+2)(x+2)(x+2)\\.\qquad \quad =(x+2)(x^2+4x+4)\\.\qquad \quad =x(x^2+4x+4)+2(x^2+4x+4)\\.\qquad \quad =x^3+4x^2+4x\quad +2x^x+8x+8\\.\qquad \quad =x^3+(4x^2+2x^2)+(4x+8x)+8\\.\qquad \quad =x^3+6x^2+12x+8$

$\dfrac{(x+2)^3-8}{x}=\dfrac{(x^3+6x^2+12x+8)-8}{x}=\dfrac{x^3+6x^2+12x}{x}\\\\\\=\dfrac{x(x^2+6x+12)}{x}=x^2+6x+12\\\\\\ \lim_{x \to 0} \quad x^2+6x+12\quad =\quad 0+0+12\quad =\quad \large\boxed{12}$