Question

Find the limit as h approaches 0 of (f(x+h)-f(x-h))/(2h).

Answer 1

any particular \(f(x)\)?

Answer 2

no

Answer 3

then it could be equal to anything, or not even exist at all

Answer 4

but perhaps you are supposed to look at it is \[\frac{1}{2}\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]

Answer 5

do you recognize the expression after the \(\frac{1}{2}\) ?

Answer 6

Yes, that would just be f'(x)/2

Answer 7

yes

Answer 8

But if you are going to do it that way, then it would be written
\[\frac{ 1 }{ 2 }[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }+\lim_{h \rightarrow 0}\frac{ f(x-h)+f(x) }{ h }]\]

Answer 9

oooh i read it wrong, sorry

Answer 10

no problem, I'm just a bit confused as to how to get a general solution to the problem

Answer 11

what you wrote is the derivative twice

Answer 12

wait...really?

Answer 13

So the second limit is also a derivative?

Answer 14

i think so, let me check more carefully hold on
yes the second is also since it is a two sided limit
\(h\) has to go to zero from both directions so there is no difference between them

Answer 15

oh no wrong again, let me pay better attension
the second is plus not minus

Answer 16

but if the function is continuous then \[\lim_{h\to 0}f(x-h)=f(x)\]

Answer 17

i better do this with pencil and paper so as not to mess you about

Answer 18

ok, I'm not told that the function is continuous, and I'm not really sure whether it would be proper to infer that, but yes, it would make the problem a whole lot easier if it was continuous

Answer 19

if it isn't then you can't say anything

Answer 20

So, would the answer just be f'(x)/2 ?

Answer 21

no just \(f'\) since you have two of them and you are dividing by 2
lets go carefully and make sure it it is clear and also correct

Answer 22

you already broke it apart by adding and subtracting \(f(x)\) to get \[\frac{ 1 }{ 2 }[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }+\lim_{h \rightarrow 0}\frac{ f(x-h)+f(x) }{ h }]\]

Answer 23

oops typo there, it should be \[\frac{ 1 }{ 2 }[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }+\lim_{h \rightarrow 0}\frac{ -f(x-h)+f(x) }{ h }]\]

Answer 24

the first part is evidently the derivative right?

Answer 25

yeah

Answer 26

to see that the second one is also replace \(h\) by \(-h\)

Answer 27

let me know if that is clear or not

Answer 28

Yeah, that makes sense, because -h would turn the second limit into the second limit. But if you replace h by -h for the second limit, wouldn't you have to do that with the first one as well?

Answer 29

no

Answer 30

taking the limits separately

Answer 31

ok, and I have another question: would replacing h with -h mean that h would have to be negative for the limit to work?

Answer 32

no h goes to zero from both directions
it is just an algebra trick to turn \[\frac{-f(x-h)+f(x)}{h}\] in to \[\frac{f(x+h)-f(x)}{h}\]

Answer 33

ok, that makes sense. Thank you for your help!

Answer 34

yw

Answer 35

sorry i screwed up so much before reading properly

Answer 36

don't worry, we all make mistakes