Give the rational function f(x)=x^4+2x^3-4x^2+8x+5/x^2+5x+6
What is the limits as x is going toward - infinity f(x)=?
I did x^4/x^2=x^2/X=X/1 so I got one as my answer.
Is this correct?
Thank you for checking my work.
as x gets large, f(x) ≈ x^4/x^2 = x^2
so f(x) -> +∞
a little confused about that, but thank you for your help and checking over my work.
as x -> ∞, the high powers grow so much faster than the lower powers that they become effectively zero, so you can ignore them.
Alternatively, divide top and bottom by x^2, That gives you
(x^2+2x-4 + 8/x + 5/x^2) / (1 + 5/x + 6/x^2)
Now, you know that 1/x and 1/x^2 both -> 0 as x ->∞
So the fraction becomes effectively
(x^2+2x-4)/1
which, of course, -> ∞ as x does.
To find the limit of the given rational function as x approaches negative infinity, you need to divide both the numerator and denominator by the highest power of x. In this case, the highest power is x^2.
Dividing both the numerator and denominator by x^2, we get:
f(x) = (x^4+2x^3-4x^2+8x+5) / (x^2+5x+6)
= (x^4/x^2 + 2x^3/x^2 - 4x^2/x^2 + 8x/x^2 + 5/x^2) / (x^2/x^2 + 5x/x^2 + 6/x^2)
= (x^2 + 2x - 4 + 8/x + 5/x^2) / (1 + 5/x + 6/x^2)
Now, as x approaches negative infinity, both 8/x and 5/x^2 approach 0. Additionally, 5/x^2 approaches 0 faster than 8/x. The terms 5/x^2 and 6/x^2 in the denominator also tend to 0. This leaves us with:
f(x) ≈ (x^2 + 2x - 4) / 1
Now, as x approaches negative infinity, the quadratic term dominates and x^2 becomes much larger than 2x and -4. Therefore, the limit of x^2 + 2x - 4 as x approaches negative infinity is positive infinity.
In conclusion, the limit as x approaches negative infinity of f(x) is positive infinity, not 1. So, your initial answer was incorrect.