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??
3 answers
We learned about the quotient rule and product rule for derivatives and thats really it. So I didnt know the formula you sent me
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
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
1 answer