Limit Questions? Check?

Question

I just want to make sure I'm doing these right.

For g(x) = 3x /(x-2)

lim g(x) as x approaches zero

The limit does not exist, right?

And for lim (x^2-4)/(x^3 +8), limit as x approaches -2,

the limit is -2/5?

Thanks.

Answer 1

lim [x → 0] 3x/(x - 2)

= 3(0)/(0 - 2)

= 0/-2

= 0

Now for the second notice that:

g(x)

= (x² - 4)/(x³ + 8)

= [(x + 2)(x - 2)]/[(x + 2)(x² - 2x + 4²)]

= (x - 2)/(x² - 2x + 4)

Therefore, we have that:

lim [x → -2] (x² - 4)/(x³ + 8)

= lim [x → -2] (x - 2)/(x² - 2x + 4)

= (-2 - 2)/((-2)² - 2(-2) + 4)

= -4/12

= -1/3

Hope this helps!

Answer 2

Sorry, 0 & -1/3...for g(x) the top gets "close" to 0 while the bottom to -2 and in the 2nd you should note that  (x^3 + 8) = (x+2)(x² - 2x +4)...thus you can cancel the factors (x+2) from top and bottom and then work the problem