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".

 

[CellCoree]

Great job on solving the problem! It seems like you used the correct method by factoring out the denominator and then plugging in the value for x. This is a common method for solving limits involving polynomials. As for a simpler way, it really depends on the level of math you are currently studying. If you are just starting to learn about limits, then this method is appropriate. However, if you are more advanced and have learned about derivatives, you could use L'Hopital's rule to solve this limit. This rule allows you to take the derivative of both the numerator and denominator and then plug in the value for x. In this case, you would have to use the power rule to take the derivative of x^5 and then evaluate it at x=2. This would give you the same answer of 80. So, while there may be alternative methods of solving this limit, the method you used is perfectly valid and appropriate for your current level of math understanding. Keep up the good work!