Calculus Problems Limits

What is the answer to the problem of calculation? LIMITS!?

the problem is … 1) Use limits to find the horizontal asymptotes for the following function f (x) = 6x ^ 2 +2 ^ 2 +3 x-4/2x X = 2. Support your answer algebraically by factoring both numerator denomnation. in addition, support your answer graphically. please help! any help will be enough Plesae need the job! THANK especially factoring, but the end is a 2x ^ 2 +3 x 2 … have to change your answer?

ok so the factor of the equation, which make normal factoring up and down. I will assume that the equal sign does not an end -. f (x) = 6x ^ 2 +2 ^ 2 +3 X-2 x-4/2x to the top of factoring (find two numbers that add up to 2 and multiply to 6 *- 4 = -24, these numbers are 6 and -4): 6 x ^ 2 + 2x – 4 6x ^ 2 + 6x – 4x – 4 6x (x +1) – 4 (x +1) (6x-4) (x +1) 2 (3x -2) (x +1) and factoring the bottom (add a 3 and multiply -4, which is 4 and -1 term. These numbers determine how to divide x): 2 x ^ 2 + 3x – 2 2x ^ 2 + 4x – x – 2 2x (x 2) – 1 (x 2) (2x-1) (x 2) then f (x) = [2 (3x-2) (x +1)] / [(2x-1) (x 2)] to find the horizontal asymptotes, we must find the limit as x tends to infinity when they are calculating the limits as x becomes very very large, general neglect constant terms. Because I mean 999999999999999999999 + 2 is close enough to 999999999999999999999. so if you take the limit as x gets larger, you get a cancellation of the infinities. f (x) = [2 (3x-2) (x +1)] / [(2x-1) (x + 2)] = (2 * 3 x * x) / (2x * x) = 6x ^ 2 / 2 x ^ 2 = 3. So the horizontal asymptote is 3. If you want to find a vertical asymptote, which is the root the denominator by setting it to 0 and solving for x. [(2x-1) (x 2)] = 0, so x = 1 / 2 x = -2 are the vertical asymptotes. Hope this helped.

Calculus Limits: Basic problems