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# Nested function limits

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Hi!

I am working through a khan academy calculus course and I am a but stumped at one of their videos. I linked it below. It seems that they are using th terms "from above" and and "from below" interchangably. If not, what is that allows someone to determine the limit of a nested function when the two conditions of the theorem are not met (condition 1: there is no limit in the nexted function condition 2: the function does not continue past the stated x-value for which we are determining the limit)?

https://youtu.be/eK8T3CcSvwE

Nov 30, 2020

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This video seems to make it more difficult than necessary but it is probably doing that on purpose.

If you get your head around difficult terminology it usually means you are learning properly, hopefully at a deeper level.

A better graph representation:

$$\displaystyle \lim _{x\rightarrow 0}\;f(g(x))\\ \text{I would just have said that } g(0)=2 \;\;so\;\;\\ =\displaystyle \lim _{x\rightarrow 0}\;f(2)\\ =0$$

Let's see what the video teacher was saying

$$\displaystyle \lim _{x\rightarrow 0}\;f(g(x))\\$$

Look at the g graph first.

I am just referring to red stuff that he talked about.  He has not explained it well.

look at g function first     (vertical axis is g, horizontal axis is x)

As x approaches 0 from the negative x side (he has called the left side the below side.)

g approaches 2 from higher g values,... from above

Now move over to the f function

If you are going to find f(g(x)) then the horizontal axis become   g(x) or g      NOT x  and the vertical axis is f(g(x)) or just f for short.

So g is approaching 2 from above  means that on the f graph, g is approaching 2 from the right side.

LaTex

\displaystyle \lim _{x\rightarrow 0}\;f(g(x))\\
\text{I would just have said that } g(0)=2 \;\;so\;\;\\
=\displaystyle \lim _{x\rightarrow 0}\;f(2)\\
=0

Dec 1, 2020
edited by Melody  Dec 1, 2020
edited by Melody  Dec 2, 2020
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Thank you Melody! It doesn't make sense YET, but I will re-read your explanation a few more times and I'm sure that it will!

WillyGolden  Dec 1, 2020
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Melody  Dec 2, 2020
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Thanks, Melody.....that was a LOT of work   !!!!

CPhill  Dec 2, 2020
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Thanks Chris,

It wasn't that much really.  Thinking it through required a bit of effort though.

I hope Willy, or any other interested person, can work through what I have said.

Melody  Dec 2, 2020