Question

proof question, I need to prove that \[f'(c)=\lim_{h\rightarrow 0} \frac{f(c+h)-f(c-h)}{2h}\]

Answer 1

So my issue is getting 2h on the bottom. Through a substitution from the definition of derivative of h=x-c, I can get the top and bottom minus the 2. Any ideas?

Answer 2

@satellite73 any thoughts?

Answer 3

\[\lim_{h\rightarrow 0} \frac{f(c+h)-f(c-h)}{2h}\]
\[\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{2h}-\lim_{h \rightarrow 0}\frac{f(c-h)-f(c)}{2h}\]

Answer 4

then I think it should be good from here right?

Answer 5

pull out the constant multiple
and you know f'(c) equals you know what

Answer 6

ahh add c subtract c. Thanks

Answer 7

oh wait, no that doesn't get me the 1/2

Answer 8

add f(c) and subtract f(c)
and if you aren't convinced that one thing is f'(c) you could substitute the -h for u

Answer 9

we cannot assume the conclusion

Answer 10

\[\frac{1}{2}f'(c)-\frac{1}{2} f'(c)\]

Answer 11

oh your right it gives you 0

Answer 12

Let f(x)=x^2
f(c)=c^2
f(c+h)=(c+h)^2=c^2+2ch+h^2
f(c-h)=(c-h)^2=c^2-2ch+h^2\[\lim_{h \rightarrow 0}\frac{c^2+2ch+h^2-c^2+2ch-h^2}{2h}=\lim_{h \rightarrow 0}\frac{2ch+2ch}{2h}=\lim_{h \rightarrow 0}(c+c)=2c\]
lol just had to make sure

Answer 13

I just cannot figure out the two... but if I compare it to the definition I know that involves h \[f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]We haven't proved that, but it is the def I know from calc... so somehow, f(x)=1/2f(x-h)

Answer 14

I get that lol

Answer 15

I didn't buy it at first either

Answer 16

I am trying to turn \[\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}\] into that

Answer 17

yeah that is what i'm trying to do now

Answer 18

So my first step was a substitution of h=x-c which turns it into \[\lim_{h\rightarrow 0}\frac{f(c+h)-f(x-h)}{h}\] Ahhh! we have to change the x-h into a c-h right!

Answer 19

\[\lim_{u \rightarrow c}\frac{f(u+2h)-f(u)}{2(c-u)}\]
hmmm...

Answer 20

so we input so we put in the c+h for the x... no that doesn't work. We just get c then

Answer 21

wouldn't yours have to be a double limit?

Answer 22

I think I made a typeo
let's see u=c-h
u+h=c
h=c-u
so c+h=c+(c-u)=2c-u

Answer 23

\[\lim_{u \rightarrow c}\frac{f(2c-u)-f(u)}{2(c-u)}\]\[\lim_{u \rightarrow c}\frac{f(2c-u)-f(u)}{2c-2u}\]

Answer 24

Like the bottom would have been cooler if we had (2c-u)-u
but we don't

Answer 25

yea... hmm

Answer 26

why do you have u=c-h?

Answer 27

was making a substitution to see if i could get it in the form\[\lim_{u \rightarrow c}\frac{f(u)-f(c)}{u-c}\]

Answer 28

or the other way around whatever \[\lim_{u \rightarrow c}\frac{f(c)-f(u)}{c-u}\]

Answer 29

\[f'(c)=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h} \\ f'(c)=\lim_{-h \rightarrow 0}\frac{f(c-h)-f(c)}{-h}=\lim_{h \rightarrow 0}\frac{f(c)-f(c-h)}{h} \\ f'(c)+f'(c)=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h}+\lim_{h \rightarrow 0}\frac{f(c)-f(c-h)}{h}\]

Answer 30

we can't go from that def because we haven't proved it :/

Answer 31

but the other one the one mentioned just a second ago?

Answer 32

I did, we used it in calc, but since we haven't proved it in this class, we are not allowed to use it without deriving it

Answer 33

\[\lim_{u \rightarrow c}\frac{f(u)-f(c)}{u-c}=f'(c)?\]

Answer 34

can we use that one?

Answer 35

what was your u again? or is it the same as x?

Answer 36

if it is the same as x,, yes we can use it

Answer 37

yeah you can think of u as x
it is just a variable

Answer 38

you had a substitution prior, so I wasn't sure if you were still using it

Answer 39

I will use x if you need me to
\[\lim_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}=f'(c)\]
let x-c=h
so we have
as x->c then h->0 \[\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h}=f'(c)\]
you just proved it now you can use it

Answer 40

meh, ok, that's what I showed earlier, only issue was the secondary x substitution on f(x-h) I was thinking it may not be alright

Answer 41

What? I don't understand?

Answer 42

We just showed that following are equivalent:
\[\lim_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h}=f'(c)\]


\[f'(c)=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h} \\ f'(c)=\lim_{-h \rightarrow 0}\frac{f(c-h)-f(c)}{-h}=\lim_{h \rightarrow 0}\frac{f(c)-f(c-h)}{h} \\ f'(c)+f'(c)=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h}+\lim_{h \rightarrow 0}\frac{f(c)-f(c-h)}{h} \]

Answer 43

well in order to get the f(c) you have to substitute in x=h+c

Answer 44

that was where I was having an algebra moment of forgetfulness, but anyways, I still don't see how we get 2h alone on the bottom

Answer 45

what is f'(c)+f'(c)

Answer 46

...ah ok I missed that equality

Answer 47

ok, I got it from here, I just have to add justification to the left limit and hope they accept it. I still do not know how to properly write up a left/right limit, since they technically don't exist in this class

Answer 48

thanks again

Answer 49

what do you mean write up a left/right limit?

Answer 50

is that a different question?

Answer 51

no, but I have to justify the left handed limit concept used to switch the top around.

Answer 52

you need to use left and right limits to prove this?

Answer 53

I'm sorry I'm having trouble understanding.

Answer 54

You used the fact that the left hand limit is equal to the limit in your step 3

Answer 55

oh i didn't do anything with left and right limits

Answer 56

you used -h, and h

Answer 57

orrrrrrr those are two separate sequences. I can spin that

Answer 58

\[f'(c)=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h} \\ h=-a \\ h->0 \text{ then } -a->0 \\ f'(c)=\lim_{-a \rightarrow 0}\frac{f(c-a)-f(c)}{-a} \ \text{ replace }a \text{ with } h \]
but if -a->0 then a->0
\[f'(c)=\lim_{a \rightarrow 0}\frac{f(c-a)-f(c)}{-a}\]

Answer 59

but you can replace the a back with h

Answer 60

also people i think usually symbolize left limit like this h->0^-

Answer 61

that is using the concept of left/right limits, but it's cool. Conceptually that is what is occurring. We can't say that \(lim{~a\rightarrow 0}=lim {-a \rightarrow 0}\)

Answer 62

but anyways thanks, I got this question from here

Answer 63

\[\lim_{x \rightarrow 5}(x+6)=5+6=11 \\ \lim_{-x \rightarrow 5}(-x+6)=-(-5)+6=5+6=11\]

Answer 64

if a->0 then so does -a

Answer 65

it is like saying if a=0
then -a=0

Answer 66

that's not the same question

Answer 67

i didn't use left and right derivatives or limits

Answer 68

for your example(the second is \(-x\rightarrow -5\), and second. You have to rigorously prove it, we can't just use it

Answer 69

no i have -x->5

Answer 70

but anyways, it doesn't matter we are arguing over nothing at the moment

Answer 71

you have to use substitutions to prove this

Answer 72

then your limt would =1

Answer 73

sorry, just your work was weird

Answer 74

you shouldn't have -(-5) it should just be 5 based off of your limit

Answer 75

\[\lim_{x \rightarrow 5}(x+6)=5+6=11 \\ \lim_{-x \rightarrow 5}(-x+6)=-(-5)+6=5+6=11 \]
Ok we have \[\lim_{x \rightarrow 5}(x+6) \text{ \let } x=-u \\ \text{ so we have } \lim_{-u \rightarrow 5}(-u+6)\]

Answer 76

though thy are logically equivalent

Answer 77

so like I said those are both equal to 11

Answer 78

yes, but the work was incorrect, which is what I look at

Answer 79

but anyways, like I said it doesn't matter at this point

Answer 80

the work isn't inccorrect

Answer 81

if -x->5 then x->-5

Answer 82

so i replaced the x with -5

Answer 83

i could have replaced -x with 5
or x with -5

Answer 84

yes, it is \[lim_{−x→5}(−x+6)=\color\red{−(−5)}+6=5+6=11\]
The step in red is not correct, based off of the limit you exchange the entire -x and turn it into a positive 5. You do not insert a negative 5 into x.

Answer 85

we just said the same thing essentially, so anyways thanks for the help, have a good night

Answer 86

no you are saying what I was incorrect
and i'm telling you it isn't

Answer 87

but anyways I think I'm done with this

Answer 88

but if you need someone else to clarify what I have said
then I will holler at @ganeshie8
maybe he will be able to prove it to you

Answer 89

\[
\begin{align}\lim_{h\to 0} \frac{f(x)-f(x-h)}{h} &= \lim_{t=-h\to 0} \frac{f(x)-f(x+t)}{-t}\\~ \\&= \lim_{t=-h\to 0} \frac{f(x+t)-f(t)}{t}\\~\\ & = f'(x)
\end{align}\]

Answer 90

oh its already there above, i have nothing else to say here :O
http://prntscr.com/52v507

Answer 91

@Kainui

Answer 92

use definition of slope :)