Question

How do you find the limit of `((2x^2-6)/(5x-x^2))` as x approaches infinity?

Answer 1

`lim_(x ->oo)((2x^2-6)/(5x-x^2)) = -2`

Explanation:

This is one way to approach this problem :

`lim_(x ->oo)((2x^2-6)/(5x-x^2))`

If you notice the highest degree of both the denominator and numerator is 2. We can divide every term by `x^2` to get

`lim_(x->oo) (((2x^2)/x^2 -6/x^2) /((5x)/x^2 -(x^2/x^2)))`

`lim_(x->oo) ((2(cancelx^2)/cancel(x^2) -6/x^2) /((5x)/x^2 -cancel(x^2/x^2)))`

`lim_(x->oo)(2-6/x^2)/(5/x-1)`
`lim_(x->oo)(2-6/(oo)^2)/(5/oo-1)`

Note: As the denominator get larger, the number will be smaller, and almost close to 0. We can stated as follow

`(2-0)/(0-1) = -2`

`lim_(x ->oo)((2x^2-6)/(5x-x^2)) = -2`

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The other method is to use L'Hopitals' Rule

If we direct substitute we will get an intermediate form

Direct Sub: `(2(oo)^2-6)/(5(oo)-(oo)^2) = -(oo)/(oo)`
We can simply differentiate numerator and denominator separately like so

`lim_(x->oo)((2x^2-6)/(5x-x^2))`

Note: Derivative of `2x^2-6 = 4x`

Derivative of `5x-x^2 = 5-2x`

We can rewrite it as : `lim_(x->oo)((2x^2-6)/(5x-x^2)) =lim_(x->oo)((4x)/(5-2x))`

by direct substitution, we get `(-oo)/(oo)` = intermediate form

We can differentiate again to get

`lim_(x->oo)((2x^2-6)/(5x-x^2)) =lim_(x->oo)((4x)/(5-2x))=lim_(x->oo) (4)/(-2) = -2`