What is the limit as x approaches infinity of #(x^2-4)/(2+x-4x^2)#?

1 Answer
Sep 28, 2014

#lim_{x to infty}{x^2-4}/{2+x-4x^2}=-1/4#

Let us look at some details.
There are a couple of methods to do this.

Method 1

#lim_{x to infty}{x^2-4}/{2+x-4x^2}#

by dividing the numerator and the denominator by #x^2#,
(#x^2# was chosen to match the leading term in the denominator.)

#=lim_{x to infty}{1-4/x^2}/{2/x^2+1/x-4}={1-0}/{0+0-4}=-1/4#

Method 2

#lim_{x to infty}{x^2-4}/{2+x-4x^2}#

by l'Hopital's Rule (#infty#/#infty#),

#=lim_{x to infty}{2x}/{1-8x}#

by l'Hopital's Rule (#infty#/#infty#) again,

#=lim_{x to infty}{2}/{-8}=2/{-8}=-1/4#

I hope that this was helpful.