as x approaches 0
the first way would be to evaluate it which doesn't work because we divide by 0
hmm, we could expand
ya, ok,
so
if we expand, we get
[tex]\frac{x^3+6x^2+12x+8-8}{x}=\frac{x^3+6x^2+12x}{x}=[/tex]
[tex]\frac{x(x^2+6x+12)}{x}=x^2+6x+12[/tex]
we can now evaluate for x=0
[tex]0^2+6(0)+12=0+0+12=12[/tex]
the value of the limit is 12
