How do you find the vertical, horizontal or slant asymptotes for #(4+x^4)/(x^2-x^4)#?

1 Answer
Apr 27, 2018

Answer:

Vertical asymptotes at #x=-1#, #x=0# and #x=1# and horizontal asymptote at #y=-1#.

Explanation:

We have vertical asymptotes for #f(x)# at #x=a#, when #lim_(x->a)f(x)=+-oo#. If #f(x)# is an algebraic function, which happens for values of #x#, for which denominator is #0#.

Here in denominator, we have #x^2-x^4=x^2(1-x^2)# and this is equal to zero when #x=0,1# or #-1#.

Hence, we have vertical asymptotes at #x=-1#, #x=0# and #x=1#.

We have horizontal asymptote when #lim_(x->+-oo)f(x)=k#, a constant. This hapens in algebraic functions, when degree of numerator is equal to that of denominator, which is the case here.

Here #lim_(x->+-oo)(4+x^4)/(x^2-x^4)#

= #lim_(x->+-oo)(4/x^4+1)/(1/x^2-1)#

= #1/(-1)=-1#

Hence we have a horizontal asymptote #y=-1#

Slant asymptote is there when degree of numerator is just one more than that of denominator. Here we do not have a slant asymptote.

graph{(4+x^4)/(x^2-x^4) [-10, 10, -10, 25]}

As an example of slant asymptote consider #(4+x^5)/(x^2-x^4)#

#lim_(x->+-oo)(4+x^5)/(x^2-x^4)#

= #lim_(x->+-oo)(4/x^4+x)/(1/x^2-1)#

= #x/(-1)=-x# and slant asymptote is #y=-x# or #x+y=0#

graph{(4+x^5)/(x^2-x^4) [-10, 10, -10, 25]}

Graphs not drawn to scale.