How Do You Solve the Limit of (x^3-2x)/(x-2) as x Approaches 2?

[skateza]

Homework Statement

lim x->2 (x^3-2x)/(x-2)

i have done this one thousand ways... factor out the x, multiply top and bottom by (x-2) and (x+2)... i have tried lots of things but i can't get it to work, any tips how to start this off?

 

[bel]

Try long division and then express the remainder term as a rational function, and then apply L'hopital's rule to it.

 

[skateza]

yeah, we definitely haven't gone that far...

 

[Gib Z]

Long division of polynomials is just like normal long division and many tutorials are on the internet. Using it we get the following result: $$\frac{x^3-2x}{x-2} = x^2 + 2x+ 2 + \frac{4}{x-2}$$.

You should check it by expanding the right hand side.

The first part, the polynomial, evaluates to 10. The fraction's denominator goes to zero, so that part goes to infinity. Hence the limit does not exist (the function approaches infinity as x approaches 2).

bel, you can't apply L'hopital's rule to the ration function because it does not fit any of the indeterminate forms. Using it leads to an incorrect result.

 

[genneth]

Homework Statement

lim x->2 (x^3-2x)/(x-2)

i have done this one thousand ways... factor out the x, multiply top and bottom by (x-2) and (x+2)... i have tried lots of things but i can't get it to work, any tips how to start this off?

The first thing you should do with a limit is to put the number in, completely ignoring the fact that it's a limit. If you get something like x/0 where x isn't 0, then no fooling about with anything will make that limit finite. So in this case, the limit is simply infinite.