How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x^2+1)/(x+1)#?

1 Answer
Oct 23, 2016

Answer:

A vertical asymptote is #x=-1#
And the oblique asymptote is #y=x-1#

Explanation:

As #f(x)=(x^2+1)/(x+1)#
Since we cannot divide by #0#
The degree of the binomial of the numerator #>#the degree of the binomial of the denominator, so we would expect an oblique asymptote
Let 's do a long division

#x^2##color(white)(aaaaaaaa)##+1##color(white)(aaaaaa)##∣##x+1#
#x^2+x##color(white)(aaaaaaaaaaaaa)##∣##x-1#
#color(white)(aaaa)##0-x+1#
#color(white)(aaaaaa)##-x-1#
#color(white)(aaaaaaaa)##0+2#

And finally we get
#f(x)=x-1+2/(x+1)#
So the oblique asymptote is #y=x-1#

#lim f(x)=-oo#
#x->-oo#

#lim f(x)=+oo#
#x->+oo#