Factoring A difference of fifths

[PhysicsAdvice]

One of the homework questions i did tonight was (x^5-32)/(x-2), where i had to find the limit when x approaches two. In order to do this i looked up the formula to use for a difference of fifths and was able to solve the question by factoring out the denominator and putting two in for x, giving me an answer of 80.

I am wondering if there is a much simpler way of solving this which i foolishly missed as we are definitely not expected to know the formula for a difference of fifths and I'm not sure it is easily derived. If anyone can spot one or argue that there isn't one that would be great! :)

 

[Hurkyl]

One of the homework questions i did tonight was (x^5-32)/(x-2), where i had to find the limit when x approaches two. In order to do this i looked up the formula to use for a difference of fifths and was able to solve the question by factoring out the denominator and putting two in for x, giving me an answer of 80.

I am wondering if there is a much simpler way of solving this which i foolishly missed as we are definitely not expected to know the formula for a difference of fifths and I'm not sure it is easily derived. If anyone can spot one or argue that there isn't one that would be great! :)

L'hôpital's rule is probably the easiest.

Doing the polynomial division is a close second, I think.

 

[PhysicsAdvice]

having not learned derivatives yet i suppose the method i indicated is the most suitable then?

 

[HallsofIvy]

It is generally true that x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1}).

That seems the best method to me. You should not have to "look it up".