Wednesday, 3 July 2019

calculus - Finding vertical asymptotes



So I am trying to find the behaviour of this function around an asymptote at x=0.



a>0



y=(xa)(x2+a)x2



I know that as x0+, y and x0, y.




I've tried dividing the top and bottom of y by x2 and ended with:



xa+axa2x2



Which does not help me as I end with . Any help would be greatly appreciated.


Answer



Expanding my comment above.





I know that as x0+, y and x0, y.




Only the first assertion is correct.



Since (xa)(x2+a)=x3ax2+axa2, we have
limx0(xa)(x2+a)x2=limx0(x3ax2+axa2)×limx01x2=a2×limx01x2=a2(+)=, for a0.



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).



enter image description here



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