Question

See 29:10 to 29:20 of this video: http://www.youtube.com/watch?v=ElCrxgHyowlet&feature=relmfu

He is going about with developing the derivative of the function ln(x). He said that that "moving through" idea can work with any continuous function. My question is: how would it work for a function like f(x) = x^3?

Could someone show me how it would be written, just like how he did it with the ln function?

Answer 1

I can't see the video.

Answer 2

If you can explain what you're looking for I can probably help.

Answer 3

Ok this is what is happening:

Answer 4

He is using the following, based on the rules of Continuous functions:
ln(a) = lim(ln u), as u approaches a

I understand how this works: you just make a direct subtitution of the value a when you evaluate the limit. But then he does this:

Answer 5

He says that that expression "ln(a) = lim(ln u), as u approaches a", can be written out like this as well:

ln(a) = ln[lim u, as u approaches a] <-- sorry if this is not clear. Just think of the "u approaches a" part as what you usually see underneath the "lim".

Answer 6

He says this idea can work for all countinuous functions; the idea of moving around the limit with the function.

Answer 7

So wait let me write it out again in mathematical form:

Answer 8

I get it.

Answer 9

\[\ln(a) = \ln[\lim_{u \rightarrow a}[u]]\]

Answer 10

The limit of the function is the function of the limit.

Answer 11

wait can you say that again with the situation above?

Answer 12

He's saying that \[\lim_{u\rightarrow a} [f(u)] = f(\lim_{u\rightarrow a}[u] )\]

Answer 13

Ok say we use the continuous function f(x) = x^3

Answer 14

how would the expression that you wrote above be written?

Answer 15

\[\lim_{x\rightarrow a}\left[ x^3\right] = \left(\lim_{x \rightarrow a} [x]\right)^3 \]

Answer 16

Which is why we're justified to just 'plug it in' if we know the function is continuous.

Answer 17

The function is that it takes a value of "x", and cubes it?

Answer 18

Correct

Answer 19

Okay so THAT is why they do direct substitution.

Answer 20

That's why direct substitution is ok for functions that are continuous at a

Answer 21

Thank you very much.

Answer 22

You are very welcome =)