# Difference: ForumMSC0003 (2 vs. 3)

#### Revision 32010-02-26 - DickFurnas

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# limit

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Multiplying by the conjugate may not work here, since we have a power less than 1/2. If you have learned l'Hopital's rule, this provides a way to solve this rather easily, and it seems to me that this is the only way to do it, at least the only straightforward way. However, if you are certain that this problem can be solved without that, we can continue trying some other methods.

-- MattGuay - 2010-02-26

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Consider a new variable w with x = w^5
You can then write the problem as (w-2)/(w^5 - 2^5)
The denominator then has a factor of (w-2) and you can consider the limit as w approaches 2.
Useful tool: When you have fractional exponents, you can often make the problem look more familiar by introducing a new variable which translates the problem into one involving integer exponents.

-- DickFurnas - 2010-02-26

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