Question

How do you find the limit `lim (2x^2-3x+2)/(x^3+5x^2-1)` as `x->oo`?

Answer 1

`lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1) = 0`

Explanation:

Note that as `x rarr oo` then `1/x,1/x^2,1/x^3 rarr 0`

We can therefore multiply numerator and denominator by `1/x^3` (the reciprocal of the largest power in the denominator) as follows:

`lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1) = lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1)*(1/x^3)/(1/x^3)`
`" " = lim_(x rarr oo) ((1/x^3)(2x^2-3x+2))/((1/x^3)(x^3+5x^2-1))`
`" " = lim_(x rarr oo) (2/x-3/x^2+2/x^3)/(1+5/x-1/x^3)`
`" " = (0-0+0)/(1+0-0)`
`" " = 0`

We can verify this result by looking at the graph of `y=(2x^2-3x+2)/(x^3+5x^2-1)`
graph{(2x^2-3x+2)/(x^3+5x^2-1) [-7, 13, -4.16, 5.84]}
and indeed it does appear that for large `x` the function is approaching a horizontal asymptote `y=0`