Question

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

Answer 1

`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.