If the limit as x approaches infinity (6x^2/200-4x-kx^2)=1/2, then k=

Question

A. 3
B. -3
C. 12
D. -12
E. -3

Answer 1

lim [ 6x^2 / (200-4x-kx^2) ] as x->infinity = 1/2

Notice that the numerator and denominator has the same highest degree of x. Therefore, we can use the L'hopital's rule. To do this, we divide both numerator and denominator by x^2:
lim [ (6x^2 / x^2) / (200-4x-kx^2)/x^2 ] as x->infinity = 1/2
lim [ 6 / (200/x^2 - 4/x - k) ] as x->infinity = 1/2
Substituting x = infinity,
= 6 / ( 0 - 0 - k ) = 1/2
= 6 / (-k) = 1/2
Therefore,
k = -12

hope this helps~ `u`

Answer 2

*Correction*
I'm sorry, the one I did is not called L'hopital's rule. That is just another technique of finding limits (dividing numerator and denominator by the variable with highest degree).
L'hopital's rule is different, for it uses separate differentiation of numerator and denominator when the answer is you're getting is indeterminate, if x in the limit is substituted. Actually, we can also use it here (assuming you've already covered derivatives in class):
derivative of 6x^2 = 12x
derivative of 200 - 4x - kx^2 = -4 - 2kx

If we substitute x = infinity, we'll still get an answer of indeterminate. Thus we use the rule again:
derivative of 12x = 12
derivative of -4 - 2kx = -2k
Rewriting,
lim 12 / -2k as x->infinity = 1/2
Substituting x = infinity,
= 12 / -2k = 1/2
= 6 / -k = 1/2
Thus, k = -12

I hope that's clear now.
Sorry about that, I think I need to get some sleep now. ^^;