Question

Question

Answer 1

\[\lim_{h \rightarrow 0} \frac{ (2+h)^{2}-4 }{ h }\]

Answer 2

Was thinking something like this \[\frac{ (2+h)(2+h) = 4+4h+h^{2}-4 }{ h }\]

Answer 3

\(\color{#000000}{ \displaystyle \lim_{h\to 0 } \frac{(2+h)^2-2^2}{h} }\)

Answer 4

So, you are using first principles of differentiation backwards.

Answer 5

how does the derivative relate to a limit?

Answer 6

This is the the first-principles definition:
\(\color{#000000}{ \displaystyle f'(x)=\lim_{h\to 0 } \frac{(x+h)-f(x)}{h} }\)

Answer 7

it is the same thing.
All rules of differentiation are some how derived from limits.

Answer 8

I can explain geometrically if I have time after we do the limit.

Answer 9

I wrote that limit incorrectly.

Answer 10

\(\color{#000000}{ \displaystyle f'(x)=\lim_{h\to 0 } \frac{f(x+h)-f(x)}{h} }\)

Answer 11

This is the way I understand it

derivative = rate of change in one variable in respect to another,
limit = what the function approaches as x get's close to a number

Answer 12

The general case:
\(\color{#000000}{ \displaystyle f'(x)=\lim_{h\to 0 } \frac{f(x+h)-f(x)}{h} }\)

So the derivative at x=a is:
\(\color{#000000}{ \displaystyle f'(a)=\lim_{h\to 0 } \frac{f(a+h)-f(a)}{h} }\)

Look at your limit.
\(\color{#000000}{ \displaystyle \lim_{h\to 0 } \frac{(2+h)^2-4}{h} }\)

And there is a clue:
\(\color{#000000}{ \displaystyle \lim_{h\to 0 } \frac{(2+h)^2-2^2}{h} }\)

Doesn't that look like?
\(\color{#000000}{ \displaystyle f'(\color{red}{2})=\lim_{h\to 0 } \frac{(\color{red}{2}+h)^2-\color{red}{2}^2}{h} }\)

where \(\color{#000000}{ \displaystyle f(x)=x^2 }\).

Answer 13

So, that limit is really the derivative of x^2 evaluated at x=2.

Answer 14

(Got disconnected, changed browsers. Apologize for delay)

Answer 15

When you are done with the limit, I can do the theory part.

Answer 16

Did you find what the limit approaches?

Answer 17

\[\frac{ 4h+h^{2} }{ h } = \frac{ h(4+h) }{ h } = 4+h \]

so according to this the limit

\[\lim_{x \rightarrow 0} = 4\]

Answer 18

so wait the limit gives you the value of the derivative?

Answer 19

or I think the limit gives you a function for the derivative?

Answer 20

\[\lim_{h \rightarrow 0} \frac{ (2+h)^{2}-4 }{ h }=\lim_{h \rightarrow 0} \frac{ 4+4h+h^2-4 }{ h }=\lim_{h \rightarrow 0} \frac{ 4h+h^2 }{ h }=\lim_{h \rightarrow 0} (4+h)=4\]
\[\lim_{h \rightarrow 0} \frac{ (2+h)^{2}-4 }{ h }=\left.\frac{d}{dx}(x^2)\right|_{x=2}=\left.2x\right|_{x=2}=4 \]

Answer 21

Note quite what you said, but i think you are somewhat close to the idea.

Answer 22

let me draw it:)