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??
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
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
To solve this problem, let's start by simplifying the expression:
∛(8+h) - 2 / h
To simplify the expression, let's rationalize the denominator first. Multiply the numerator and denominator by the conjugate of the expression in the denominator:
[∛(8+h) - 2] / h * [(∛(8+h) - 2)/(∛(8+h) - 2)]
This simplifies to:
[∛(8+h) - 2] * [(∛(8+h))^2 + (∛(8+h)) * 2 + 2^2] / (h * (∛(8+h) - 2))
Now we can expand and simplify further:
= [(∛(8+h))^3 - 2^3] / (h * (∛(8+h) - 2))
= [(8 + h) - 8] / (h * (∛(8+h) - 2))
= h / (h * (∛(8+h) - 2))
Notice that h simplifies out in the numerator and denominator:
= 1 / (∛(8+h) - 2)
Now that we have simplified the expression, we can take the limit as h approaches 0. Substituting h = 0 into the expression, we get:
= 1 / (∛(8+0) - 2)
= 1 / (∛8 - 2)
= 1 / (2 - 2)
As the denominator becomes 0, the expression becomes undefined.
Therefore, there is no solution to this limit as h approaches 0.