Question

I have a midterm tomorrow someone please help me study!!!

Answer 1

Hey squirrel :)
Do you remember the `other` form of the limit definition of the derivative?
I think it looks something like this:\[\Large\rm f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]

Answer 2

\[\Large\rm f'(\color{orangered}{3})=\lim_{x\to \color{orangered}{3}}\frac{f(x)-f(\color{orangered}{3})}{x-\color{orangered}{3}}\]

Answer 3

Remember this limit definition form?
It doesn't come up as often, so it's easy to forget about it.

Answer 4

the f prime?

Answer 5

whut? :U

Answer 6

im a little confused about it because i think that that can be used to calculate the slope of a tangent linear x=a, velocity at t=a, and rate of change at x=a

Answer 7

Those all mean the same thing.

Answer 8

so do all those use that same equation?

Answer 9

You're not supposed to think that hard on a question like this.
They just want you to compare `the given limit` to the `definition limit for a derivative`.

Answer 10

And match up the pieces.

Answer 11

i have two one that says f(x)-f(a)/h or
f(a+h)-f(a)/h

Answer 12

so i would just basically plug in the three into the equation?

Answer 13

Hmm, no they should be these two equations:\[\Large\rm f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]and\[\Large\rm f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]And we're using the second one.
What you have written down doesn't look like this though? :o

Answer 14

oh yea those my bad

Answer 15

yea idk how to use the equation function on here yet

Answer 16

is there a reason to why we are using the second one or did you just choose it?

Answer 17

Here is something to try and remember:

When you have a setup where \(\Large\rm h\to0\), under the limit,
you're using the first form for comparison.

When you have a setup where \(\Large\rm x\to\) a number, under the limit,
you're using the second form for comparison.

Answer 18

\[\Large\rm \lim_{\color{orangered}{x\to3}}\frac{\frac{1}{x+1}-\frac{1}{4}}{x-3}\]This orange part here is letting us know that we should compare this to the second form of the limit definition of the derivative.

Answer 19

ohh ok that is very helpful and it makes it easier to remember

Answer 20

\[\Large\rm \lim_{\color{orangered}{x\to3}}\frac{\frac{1}{x+1}-\frac{1}{4}}{x-3}\]So what can we say about \(\Large\rm a\) if we compare these?\[\Large\rm \lim_{\color{orangered}{x\to a}}\frac{f(x)-f(a)}{x-a}\]

Answer 21

so that is why we use the second one right?

Answer 22

do both equations have different names to them?

Answer 23

Hmmm I don't think so :(

\[\Large\rm \frac{f(x+h)-f(x)}{h}\]This one is called the difference quotient.
I'm not really sure what you would call the other one...
They're both doing the same thing, they're calculating the slope of a `secant` line.
The limit turns this process into finding the slope of a `tangent` line by letting the two points get closer and closer together.

Blahhh too much info.
Brain overload.

Answer 24

Let's just keep things simple for now >.<

Answer 25

Look back at what I posted, what can you say about \(\Large\rm a\) ?
Compare the orange parts...

Answer 26

x is approaching a...
x is approaching 3...

what... can we.... say... about..... .... ... ... a ?

Answer 27

comeon squirrel, you can do this :O
connect the dots

Answer 28

is 3 the point of tangency or no?

Answer 29

I don't understand why you keep dodging my question :(

Answer 30

Yes.... a=3.

Answer 31

no i was trying to answer it there but idk? sorry

Answer 32

im really confused right now

Answer 33

x is approaching 3...
x is approaching a...

so a is 3.

Answer 34

so we use 3 for f(a)?

Answer 35

and plug it in to the second equation?

Answer 36

No plug, not yet.
We need to identify another piece before we can do anything.

Answer 37

ok

Answer 38

\[\Large\rm \lim_{x\to3}\frac{\color{royalblue}{\frac{1}{x+1}}-\frac{1}{4}}{x-3}\]How should we figure out what our function f(x) is?\[\Large\rm \lim_{x\to a}\frac{\color{royalblue}{f(x)}-f(a)}{x-a}\]
Notice... the color. -_- hehe

Answer 39

we plug in 3 into that?

Answer 40

the blue part?

Answer 41

No, this has nothing to do with the 3.
We're trying to identify our function f(x) from this mess.

We're comparing two things.

Answer 42

so its just equal to the top blue part?

Answer 43

Yes. We're just matching up the pieces right now.
It's like this silly drawing here.

We have some standard model which tells us where each piece is.
And then if I give you something specific to work with like the picture on the right, you can accurately tell me that his eyes are triangular, because they're in that same location.

Nothing more than that.