Help with a calculus limit question?

Question

Find the value of k if the function
h(x)=((kx-3)(x-2))/
((7x+9)(2x+5))

has a horizontal asymptote at y=2

Answer 1

ok, multiply everything out. so we have.
h(x) = (kxx - 2kx - 3x + 6) / (14xx + 35x + 18x + 45)

now, when we're talking about horizontal asymptotes, we're worried about the leading coefficients of the term with the highest degree. in this case, the term with the highest degree in both the numerator and deonominator is xx or x^2. the leading coefficient on top is k and the leading coefficient on the bottom is 14. so, in order to have a horizontal asymptote at y = 2
k / 14 = 2
cause as x goes to infinity, all the other terms don't really matter and only xx will dominate. so, the leading coefficients of the xx will determine the horizontal asymptote.
multiply both sides by 14 and you get k = 28. graph on you're calculator just to make sure. =)

Answer 2

The numerator and the denominator are of the same degree.
So the limit as x=> infinitity is the division of the coeficients of the terms of highest degree This limit is k/14 =2 so k=28.
You wrote "at y=2" but it should say "asymptote of equation y=2"

Answer 3

[d/dx ( (kx-3) ( x-2)] / [ d/dx ( 7x +9)(2x +5)] =

[ 2kx - 2k -3]/[ 28x +51]

d/dx [ 2kx - 2k -3]/ d/dx [ 28x +51] =

2k / 28 = 2

K = 56