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

Answer 1

c does

Answer 2

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

Answer 3

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

Answer 4

2x^2/x^2 = 2

Answer 5

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

Answer 6

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

Answer 7

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