Question

lim x->0 ((1=x)^3-1/x)

Answer 1

1=x means 1+x?

Answer 2

crap! yeah

Answer 3

\[(1+x)^3-\frac{1}{x}\]so it looks like

Answer 4

what?

Answer 5

is that how the equation is written?

Answer 6

no it is written (1+x)^3-1 / x\[\left( 1+x \right)^{3} -1\div x\]

Answer 7

\[\frac{(1+x)^3-1}{x}\]so

Answer 8

correct

Answer 9

ok

Answer 10

let me type it out

Answer 11

\[\lim_{x \rightarrow 0} \frac{(x+1-1)[(x+1)^2+(x+1)+1]}{x}\]

Answer 12

That's because of a difference of 2 cubes

Answer 13

\[\lim_{x \rightarrow 0} \frac{x(x^2+2x+1+x+2)}{x}=\lim_{x \rightarrow 0} x^2+3x+3\]

Answer 14

plug in 0 for x from here, there's your solution

Answer 15

Understand?

Answer 16

The limit is 3

Answer 17

got you. Thank you!