Question

Question #59e27

Answer 1

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`

Answer 2

`-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`