Limit of (2 + h)^3 - 8 / h as h approaches 0

[@TreeCube]

There's actually a shortcut for that expression on the numerator. Notice that a³-b³ = (a-b)(a²+ab+b²), (this is just a formula you can remember), so: (2+h)³-8 = (2+h)³-2³ = [(2+h)-(2)][(2+h)²+(2+h)(2)+(2)²] = h[(2+h)²+(2+h)(2)+4] Then you can cancel the h's and get: lim h->0 (2+h)²+(2+h)(2)+4 Then plug in h = 0: = (2+0)²+(2+0)(2)+4 = 12

[@MasterWuMathematics]

Excellent work!

[@mcpcvn5050]

I'm Vietnamese and I really like your explanations, it's very easy to understand

[@壓力]

somebody tell me how to solve this problem please🙏 especially part(A)，I have no idea Let f : (-∞, ∞)--> (-∞, ∞) be a differentiable function such that f has a local minimum value f(−1) = -1, and the graph of y = f(x) contains an inflection point (0, 0), with a slant asymptote y = x + 1. (A)Prove that such a function ƒ exists by constructing a possible function. (B)Prove that there exists some a < 0 such that f'(a) = 1. (C)Show that there exists some b> 0 such that f'(b) > 1.

[@georgegonis3643]

This video was very helpful. Why does 8 cancel out?! That is my only question

[@MasterWuMathematics]

There's a +8 at the front, and a -8 at the back of the numerator... so 8 - 8 = 0

[@rando2000z]

What is h ?

[@SuperYoonHo]

Good video sir!