For which of these functions f(x) does the limit f(x) = 2 as x - -∞?
a) (x-2)/(3x-5)
b) (2x)/radical(x-2)
c) (2x^2-6x+1)/(1+x^2)
d) (2x-1)/(x^2+1)
e) none of the above

Question

For which of these functions f(x) does the limit f(x) = 2 as x -> -∞?
a) (x-2)/(3x-5)
b) (2x)/radical(x-2)
c) (2x^2-6x+1)/(1+x^2)
d) (2x-1)/(x^2+1)
e) none of the above

anonymous · Answer

c does

anonymous · Answer

since numerator and denominator both have degree 2, take the ratio  of the leading coefficients and see that you get 2 as a limit

across · Answer

I will say C:$$\lim_{x\to\infty}\frac{2x^2-6x+1}{1+x^2}=\lim_{x\to\infty}\frac{4x-6}{2x}=\lim_{x\to\infty}\frac{4}{2}=2.$$

hero · Answer

2x^2/x^2 = 2

anonymous · Answer

Thanks so much l'hopital's rule makes so much sense O.O

hero · Answer

You didn't need to use l'hopital rule for this

anonymous · Answer

It makes the most sense the way across used it though, it just seems easier, then remembering the exponent rules