Asked by Nick
How do I evaluate lim as h->0 [5((1/2)+h)^4 - 5(1/2)^4]/h
Answers
Answered by
Nick
Also , another similar question that I am having trouble with is lim as x->0 (tan^3 (2x))/(x^3)
Answered by
Damon
binomial expansion of (.5+h)^4
1* .5^4 + 4* .5^3 h + 6* .5^2 h^2 etc
5 [ .5^4 + 4* .5^3 h + 6* .5^2 h^2 + 4 .5 h^3 + h^4 -.5^4 ] / h
only term left when h--> 0 is
5 [ 4*.5^3 ]
1* .5^4 + 4* .5^3 h + 6* .5^2 h^2 etc
5 [ .5^4 + 4* .5^3 h + 6* .5^2 h^2 + 4 .5 h^3 + h^4 -.5^4 ] / h
only term left when h--> 0 is
5 [ 4*.5^3 ]
Answered by
Reiny
by definition
dy/dx = lim (f(x+h) - f(x))/h as h ---> 0
if you look at your expression carefully, you can see that they are attempting to find dy/dx of f(x) = 5x^4 when x=1/2
so dy/dx = 20x^3, which at x= 1/2 = 20/8 = 5/2
just like in Damon's answer.
dy/dx = lim (f(x+h) - f(x))/h as h ---> 0
if you look at your expression carefully, you can see that they are attempting to find dy/dx of f(x) = 5x^4 when x=1/2
so dy/dx = 20x^3, which at x= 1/2 = 20/8 = 5/2
just like in Damon's answer.
Answered by
Damon
I figured maybe the derivative is in the next chapter :)
Answered by
Nick
Thanks for the help, I know what to do now. :D
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