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

In this video, we use algebraic manipulation to find the limit as h approaches 0 of...

(2 + h)^3 - 8 / h.

If we try to find this limit as it is written, we will run into a case of 0/0, which is indeterminate.

#PreCalculus #Limits #Algebra

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Please ask me a maths question by commenting below and I will try to help you in future videos.

[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
=
= 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

[mcpcvn]

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.

[georgegonis]

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