Calculating the Limit as h Approaches 0: Power Rule Example and Explanation

[sakebu]

Homework Statement

limit as h approaches 0

Homework Equations

lim h→0 [ 2(x+h)^5 -5(x+h)^3 - 2x^5 + 5x^3 ] / h

The Attempt at a Solution

People have told me to use the power rule and gave me an answer of 10 x^4 + 15 x^2 but that doesn't seem to be right...

 

[Mark44]

Homework Statement

limit as h approaches 0

Homework Equations

lim h→0 [ 2(x+h)^5 -5(x+h)^3 - 2x^5 + 5x^3 ] / h

The Attempt at a Solution

People have told me to use the power rule and gave me an answer of 10 x^4 + 15 x^2 but that doesn't seem to be right...

The answer they gave you is a little off. Assuming that your function is f(x) = 2x⁵ - 5x³, then f'(x) = 10x⁴ -[/color] 15x².

If the problem is to find the derivative using the limit definition of the derivative, then your friends' advice of using the power rule is also incorrect. To evaluate the limit you show, expand the first two terms. You should find that some terms drop out, and you can then take the limit.

 

[SammyS]

In general: $a^{n+1}-b^{n+1}=(a-b)(a^{n}+a^{n-1}b+a^{n-1}b^{2}+\dots+a^{n-k}b^{k}+\dots+a^{2}b^{n+2}+a\,b^{n-1}+b^{n})$

So in specific, if n = 4, $a^{5}-b^{5}=(a-b)(a^{4}+a^{3}b+a^{2}b^{2}+a\,b^{3}+b^{4})$

If n = 2, $a^{3}-b^{3}=(a-b)(a^{2}+a\,b+b^{2})$

Now, let a = x+h, and b = x.

 

[sakebu]

Thanks. Once I found the binomial theorem: http://en.wikipedia.org/wiki/Binomial_theorem , It was so easy!