How do you find the asymptotes for #f(x) = (x^2) / (x^2 + 1)#?

1 Answer
Jan 22, 2016

Answer:

#cancel(EE)# vertical asymptotes
#y=1# horizontal asymptote #x rarr +-oo#

Explanation:

Given:

#f(x)=g(h)/(h(x))#

possible asymptotes are #x=x_i# with #x_i# values that zero the denominator #h(x)=0#.

Indeed, the existence condition of #f(x) in RR# is #h(x)!=0#.

Therefore,

#h(x)=0 => x^2+1=0=> x^2=-1=>x_1,x_2 in CC#

#h(x)# hasn't any zeros #in RR#

#:.f(x): RR rarr RR, AA x in RR#

We could looking for possible horizontal asymptotes:

#y=L# is a horizontal asymptotes when:

#lim_(x rarr +-oo)f(x)=L#

#lim_(x rarr +-oo)f(x)=lim_(x rarr +-oo) x^2/(x^2+1)~~cancel(x^2)/cancel(x^2)=1#

#:.y=1# horizontal asymptote #x rarr +-oo#

graph{x^2/(x^2+1) [-6.244, 6.243, -3.12, 3.123]}