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# Evaluate the limit `lim_(x->oo)(2x^3-7x^2-8x)/(9-9x-6x^3)`

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Evaluate the limit `lim_(x->oo)(2x^3-7x^2-8x)/(9-9x-6x^3)`

Posted by jmg5639 on January 23, 2013 at 2:39 AM via web and tagged with calculus, infinity, limits, math

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The limit `lim_(x->oo)(2x^3-7x^2-8x)/(9-9x-6x^3)` has to be determined. If `x->oo` , `(1/x)->0`

`lim_(x->0)(2(1/x)^3-7(1/x)^2-8/x)/(9-9/x-6*(1/x)^3)`

=> `lim_(x->0)(2(1/x)^3-7*x(1/x)^3-8*x^3/x^3)/(9*x^3/x^3-9*x^2/x^3-6*(1/x)^3)`

=> `lim_(x->0)(2-7*x-8*x^3)/(9*x^3-9*x^2-6)`

substituting x = 0

=> `2/-6`

=> `-1/3`

The limit `lim_(x->oo)(2x^3-7x^2-8x)/(9-9x-6x^3) = -1/3`

Posted by justaguide on January 23, 2013 at 3:09 AM (Answer #1)

A very similar method that might be a little bit easier to understand is to keep the `x->oo` and divide the numerator and denominator by the highest degree variable in the denominator, which in this case is `x^3` . Therefore:

`lim_(x->oo)(2x^3-7x^2-8x)/(9-9x-6x^3)` `=> lim_(x->oo) (2x^3/x^3-7x^2/x^3-8x/x^3)/(9/x^3-9x/x^3-6x^3/x^3)`

` ` Canceling out the ` `x's.

`=> lim_(x->oo) (2-7/x-8/x^2)/(9/x^3-9/x^2-6) `

Keep in mind:

`(Any Number)/oo -> 0`

So substituting `oo` for `x` :

`lim_(x->oo) (2-7/x-8/x^2)/(9/x^3-9/x^2-6) -> (2-0-0)/(0-0-6)`

`rArr` `2/-6` = `-1/3` .

Again, this method is very similar to the above response, and you get the same solution to the problem, but this method might just be a bit easier to understand.

The Answer: `lim_(x->oo) (2x^3-7x^2-8x)/(9-9x-6x^3) = -1/3`

Posted by Wilson2014 on February 20, 2013 at 6:53 PM (Answer #2)