Asked by Anonymous
limit as h approaches 0
∛(8+h)-2)/h = 1/12
The way I was shown was not something we learned so I was wondering if there was another way to do this problem??
∛(8+h)-2)/h = 1/12
The way I was shown was not something we learned so I was wondering if there was another way to do this problem??
Answers
Answered by
Anonymous
We learned about the quotient rule and product rule for derivatives and thats really it. So I didnt know the formula you sent me
Answered by
Steve
Odd. The formula I used was from Algebra I. Don't forget your old classes now that you're taking calculus.
Surely you recognize the problem as finding the derivative of
y = ∛x at x=8
y = x^(1/3)
y' = 1/3 x^(-2/3)
y'(8) = 1/8 * 8^(-2/3) = 1/3 * 1/4 = 1/12
Surely you recognize the problem as finding the derivative of
y = ∛x at x=8
y = x^(1/3)
y' = 1/3 x^(-2/3)
y'(8) = 1/8 * 8^(-2/3) = 1/3 * 1/4 = 1/12
Answered by
Reiny
How about this:
let ∛(8+h) = k
then 8+h = k^3
and h = k^3 - 8
also , as h ---> 0, k ---->2
so your problem of Lim (∛(8+h) - 2)/h, as h ---> 0
becomes Lim (k-2)/(k^3 - 8), as k ---> 2
= lim (k-2)/( (k-2)(k^2 + 2k + 4) ), as k---> 2
= lim 1/(k^2 + 2k + 4) as k---> 2
= 1/(4+4+4) = 1/12
let ∛(8+h) = k
then 8+h = k^3
and h = k^3 - 8
also , as h ---> 0, k ---->2
so your problem of Lim (∛(8+h) - 2)/h, as h ---> 0
becomes Lim (k-2)/(k^3 - 8), as k ---> 2
= lim (k-2)/( (k-2)(k^2 + 2k + 4) ), as k---> 2
= lim 1/(k^2 + 2k + 4) as k---> 2
= 1/(4+4+4) = 1/12