So I am trying to find the behaviour of this function around an asymptote at x=0.
a>0
y=(x−a)(x2+a)x2
I know that as x→0+, y→−∞ and x→0−, y→−∞.
I've tried dividing the top and bottom of y by x2 and ended with:
x−a+ax−a2x2
Which does not help me as I end with ∞−∞. Any help would be greatly appreciated.
Answer
Expanding my comment above.
I know that as x→0+, y→−∞ and x→0−, y→−∞.
Only the first assertion is correct.
Since (x−a)(x2+a)=x3−ax2+ax−a2, we have
limx→0(x−a)(x2+a)x2=limx→0(x3−ax2+ax−a2)×limx→01x2=−a2×limx→01x2=−a2(+∞)=−∞, for a≠0.
So as x approaches 0, either from the left or from the right, y→−∞. The limit does not depend on the sign of a.
Plot of y for a=1 (black) and a=−1 (green).
No comments:
Post a Comment