1. ## Limits part 1?

Can anyone help me with these two limit without using l hospital rule which I do not know.

1. As x approaches zero

(x+2)^2-4/(x)

2. As x approaches -2

(x^3+8)/(x^2-4) This is what I did

x(x^2+8)/(x+2)(x-2) Can anyone help me continue.

2. ## Re: Limits part 1?

Originally Posted by homeylova223
Can anyone help me with these two limit without using l hospital rule which I do not know.

1. As x approaches zero

(x+2)^2-4/(x)

$\displaystyle \lim_{x\to 0} \frac{(x+2)^2-4}{x}$ or $\displaystyle \lim_{x\to 0} (x+2)^2-\frac{4}{x}$ ?

3. ## Re: Limits part 1?

Originally Posted by homeylova223

x(x^2+8)/(x+2)(x-2) Can anyone help me continue.
$\displaystyle x^3+8 \neq x(x^2+8)$ , do you know why?

Consider $\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$

4. ## Re: Limits part 1?

Yes I see but how would I proceed?

5. ## Re: Limits part 1?

Originally Posted by homeylova223
Yes I see but how would I proceed?
factor $x^3+8$ correctly, then cancel the common factor in the numerator and denominator ... finally, sub in x = -2