Question #59e27

2 Answers
Feb 13, 2017

Recursively use L'Hôpital's rule

Explanation:

Because the expression evaluated at oo results in the indeterminate form -oo/oo, one should recursively use L'Hôpital's rule :

Compute the derivative of the numerator:

(d(5x^3+2x+3))/dx=15x^2+2

Compute the derivative of the denominator:

(d(4-x^2-2x^3))/dx=-2x-6x^2

Check the limit of the new expression:

lim_(xtooo)(15x^2+2)/(-2x-6x^2)

Evaluation at oo is still -oo/oo. Apply the rule a second time:

Compute the derivative of the numerator:

(d(15x^2+2))/dx=30x

Compute the derivative of the denominator:

(d(-2x-6x^2))/dx=-2-12x

Check the limit of the new expression:

lim_(xtooo)(30x)/(-2-12x)

Still -oo/oo

One more time:

Compute the derivative of the numerator:

(d(30x))/dx=30

Compute the derivative of the denominator:

(d(-2-12x))/dx=-12

Check the limit of the new expression:

lim_(xtooo)(30)/(-12) = -5/2

Therefore, the limit of the original expression goes to the same value:

lim_(xtooo)(5x^3+2x-3)/(4-x^2-2x^3) = -5/2

Feb 14, 2017

-5/2

Explanation:

Factor out of the numerator and denominator the greatest power of x in the denominator, and then reduce. (Or, divide numerator and denominator by the greatest power of x in the denominator.)

Use lim_(xrarroo)c/x^n = 0 for all c and all positive n.

lim_(xrarroo) (5x^3+2x-1)/(4-x^2-2x^3) = lim_(xrarroo) (x^3(5+2/x^2-1/x^3))/(x^3(4/x^3-1/x-2))

= lim_(xrarroo) (5+2/x^2-1/x^3)/(4/x^3-1/x-2)

= (5+0-0)/(0-0-2) = -5/2

Note

The above is the long explanation. The short one is look at just the dominating terms -- the greatest powers of x-- in the numerator and denominator. At infinity, the quotient behaves like the ratio of those greatest powers.

Examples

lim_(xrarroo)(3x^5+7x^2-19)/(4x^5-2x^3+3x-2) = lim_(xrarroo)(3x^5)/(4x^5) = lim_(xrarroo)3/4 = 3/4

lim_(xrarroo)(7x^2-5x+7)/(2x^3+5x+8) = lim_(xrarroo)(7x^2)/(2x^3) = lim_(xrarroo)7/(2x) = 0

lim_(xrarroo)(x^4+3x-9)/(6x+2) = lim_(xrarroo)(x^4)/(6x) = lim_(xrarroo)x^3/6 = oo