Question

How do you evaluate the limit `(3x^4-x^2+5)/(10-2x^4)` as x approaches `oo`?

Answer 1

`lim_(x rarr oo) (3x^4-x^2+5)/(10-2x^4) = -3/2`

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^4` (the reciprocal of the largest power in the denominator) as follows:

`lim_(x rarr oo) (3x^4-x^2+5)/(10-2x^4) = lim_(x rarr oo) (3x^4-x^2+5)/(10-2x^4)*(1/x^4)/(1/x^4)`
`" " = lim_(x rarr oo) ((1/x^4)(3x^4-x^2+5))/((1/x^4)(10-2x^4))`
`" " = lim_(x rarr oo) (3-1/x^2+5/x^4)/(10/x^4-2)`
`" " = (3-0+0)/(0-2)`
`" " = -3/2`

We can verify this result by looking at the graph of `y=(3x^4-x^2+5)/(10-2x^4)`
graph{(3x^4-x^2+5)/(10-2x^4) [-10, 10, -5, 5]}
and indeed it does appear that for large `x` the function is approaching a horizontal asymptote `y=-3/2`