- #1

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strictly algebraicly...(probably spelled that wrong)

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- Thread starter daveed
- Start date

- #1

- 138

- 0

strictly algebraicly...(probably spelled that wrong)

- #2

- 138

- 0

i get it to (5( (x)^2/5 - (x+h)^2/5) )/((x^2+xh)^2/5)

but cant get rid of the radicals... =/

but cant get rid of the radicals... =/

- #3

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daveed said:

strictly algebraicly...(probably spelled that wrong)

After about an hour (I was about to just give up) of messing around I had an idea. I needed to know if I could factor [tex]x^5 - y^5[/tex] and it turns out that I can!

For your reference

[tex]x^5-y^5 = (x-y)(x^4 + x^3y + x^2y^2 +xy^3 + y^4)[/tex]

what you have is

[tex]

5\lim_{h\rightarrow 0} \frac {x^\frac {2} {5} - (x+h)^\frac {2} {5}} {h(x^2 + 2hx)^\frac {2} {5}} [/tex]

Now the trick to getting rid of that h in the denominator is to figure out how you could go about adding the two parts of the numerator.

Start by noticing that the numerator is a difference of squares

[tex]

x^\frac {2} {5} - (x+h)^\frac {2} {5}

[/tex]

Which of course you can factor out to

[tex]

(x^\frac {1} {5} + (x+h)^\frac {1} {5})(x^\frac {1} {5} - (x+h)^\frac {1} {5})

[/tex]

Now all you have got to do is figure out how to get ride of that nasty little [tex]h[/tex] in the denominator. That is not so easy unless you multiply the entire quotient by 1. That is, multiply by

[tex]

\frac {x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}} {x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}}

[/tex]

Which, if you will recall from our earlier results, we can easily multiply the [tex] (x^\frac {1} {5} - (x+h)^\frac {1} {5})[/tex] with the [tex]x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}[/tex] and that will get rid of the fifth root and thus will give us a nice and neat numerator from which you can factor out the h. If you have gotten this far the rest is cake.

Regards

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