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

1 Answer
Jan 27, 2016

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