How to Calculate a Limit Using Cauchy's Mean Value Theorem?

  • Thread starter transgalactic
  • Start date
In summary: I almost said what you wrote but I don't know what you wrote!) It is not at all clear to me that, even with the additional information, the limit exists. You can easily construct examples where, even though f and g are differentiable at 0, f'(x) and g'(x) are not differentiable at 0. I think you are going to have to look at the definition of the derivative and see if you can prove that the limit exists. I would start by looking at the Weierstrass definition although knowing the "mean value theorem" might be useful.f(x) and g(x) are differentiable on 0f(0)=g(0)=0
  • #36
You aren't telling me anything you didn't say in post 29, and that last bit, that the derivatives are zero, is wrong.
 
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  • #37
sorry i ment to write the definition of continuity
[tex]
\lim_{x->0^+}f(x)=\lim_{x->0^-}f(x)=f(0)=0
[/tex]??
 
  • #38
Stop with the "??" stuff, please. And please answer the question raised in post #34.
 
  • #39
i don't know
i think that if the derivative exist at x=0
and because i was told that f(0)=g(0)=0
then the function around zero would have values close to 0.
 
  • #40
Say that mathematically.
 
  • #41
[tex]
\lim_{x->0^+}f(x)=\lim_{x->0^-}f(x)=f(0)=0

[/tex]
 
Last edited:
  • #42
What is the value of f(x) for x near zero?
 
  • #43
the value of f(x) near zero is close to 0
i can't imagine anything else
 
  • #44
What does [tex]f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x}[/tex] tell you about f(x) when x is close to but not equal to zero?
 
  • #45
it tells me that the slope of f(x) in the ever closing interval near point 0
(from x to 0)
gets closser to the value of f'(0)
 
  • #46
I did not ask for that and you specifically do not know that. In post #22 you said "f(x) and g(x) are not necessarily differentiable around 0".

I asked you to tell me about f(x) near 0. What is it, approximately?
 
  • #47
transgalactic said:
is that ok??
[tex]
\lim_{x->0}\frac{f(x)-f(0)}{x}=const\\
[/tex]
[tex]
\lim_{x->0}\frac{g(x)-g(0)}{x}=const
[/tex]
transgalactic said:
it tells me that the values from the right and left sides little by little become
closer to the value at f(0)

differential is also continues
[tex]
\lim_{x->0^+}\frac{f(x)-f(0)}{x}=\lim_{x->0^-}\frac{f(x)-f(0)}{x}=f(0)=0
[/tex]
??

hey guys! why is everybody going round in circles?

transgalactic, you're doing your usual problem of not quite writing what you mean …

you meant [tex]
\lim_{x->0^+}\frac{f(x)-f(0)}{x}=\lim_{x->0^-}\frac{f(x)-f(0)}{x}[/tex] = f'(0) = 0 :rolleyes:

(btw, the differential is not necessarily continuous … as you pointed out, we don't even know that it exists except at x = 0 … the equation above is the definition of f'(0), isn't it? :smile:)

hmm … where had we got to? :confused:

oh yes … we'd found that if f was differentiable over a neighbourhood, then we could use l'Hôpital's rule (twice) to get (g'(0)2 - f'(0)2)/2

but all we know about f' and g' is that they exist (and are 0) at x = 0, and they may not even exist anywhere else

but we can be pretty confident that the answer is still (g'(0)2 - f'(0)2)/2 … so let's set about proving it

let's remind ourselve of the original question:
f(x) and g(x) are differentiable on 0
f(0)=g(0)=0

calculate
[tex]\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} [/tex]

and let's rewrite that as

[tex]\lim _{x->0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}[/tex]

(we're doing that because (cos(0) - 1) is 0, and so (cos - 1) will be much more convenient than cos in a moment)

and then just consider half of it …

[tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex]

what expansion or approximation do you know for cos(f(x)) - 1 when x -> 0? :smile:
 
  • #48
the expansion for cos x around 0 is
[tex]
cosx=1+0-\frac{x^2}{2!}
[/tex]
but we need to substitute f(x) instead of x in cos x
??

and i was told specifically that the function not necessarily differentiable around 0 .
so on what basis e use lhopital law(twice)?
regard D.H question:
i don't know what is the value of f(x) around zero
i ran out of options.
 
  • #49
transgalactic said:
the expansion for cos x around 0 is
[tex]cosx=1+0-\frac{x^2}{2!}[/tex]
but we need to substitute f(x) instead of x in cos x …

(why did you write +0? :confused:

did you mean O(x4)?)
Yes … you do need to substitute f(x) … so it's
[tex]cos(f(x))=1\ -\ \frac{(f(x))^2}{2!}[/tex]
… and i was told specifically that the function not necessarily differentiable around 0 .

ah, but we haven't used f'

that equation only uses f.

ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ? :smile:
 
  • #50
i got this expression but i have x in the denominator
so its not defined when x->0 because the numenator goes to 0 too.
[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}
[/tex]
 
  • #51
So what is f(x) near zero?

Tiny-tim, please do not give this away.
 
  • #52
transgalactic said:
i got this expression but i have x in the denominator
so its not defined when x->0 because the numenator goes to 0 too.
[tex]\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}[/tex]

ok, rewrite that as [tex]\frac{-1}{2}\,\lim _{x->0}\left(\frac{f(x) }{x}\right)^2[/tex]

and since the product of the limits is the limit of the product, that equals

[tex]\frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2[/tex]

which = … ? :smile:
 
  • #53
i don't know what's the value of the limit

f(x) goes goes to f(0) but not equals f(0) so i don't know what's the value of the
numerator.
and i got 0 in the denominator

so
?
 
  • #54
Difference quotient for f'(0).
 
Last edited by a moderator:
  • #55
I think what Dick is getting at is, what is the definition of f'(0)?
 
  • #56
i don't know what is the value of f'(0)
i know that f(0)=0

[tex]
f'(x)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}

[/tex]
this is the definition of the derivative
i don't know how to continue

you said also "Difference quotient" so i used
[tex]
f'(x)=\lim _{h->0}\frac{f(x+h)-f(x)}{h}
[/tex]
but i don't have any values for it.
 
Last edited:
  • #57
D H said:
So what is f(x) near zero?

Tiny-tim, please do not give this away.
i gave every option i can think of.
i don't know.
 
  • #58
transgalactic said:
i don't know what is the value of f'(0)
i know that f(0)=0

[tex]f'(x)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}[/tex]
this is the definition of the derivative
i don't know how to continue

transgalactic, that isn't the definition of f'(x), it's the definition of f'(0) (using x instead of the more usual h, and since f(0) = 0):

[tex]f'(0)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}=\lim _{x->0}\frac{f(x)}{x}[/tex]

ok, so now you have …
[tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}\ =[/tex]
transgalactic said:
[tex]\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}[/tex]
[tex]=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2[/tex]

= … ? :smile:
 
  • #59
ok i am doing that as a shot in the dark
inspite of the fact the f(x->0) differs f(0)

and i put the given f(0)=0
[tex]
\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{0 }{0}\right)^2=
[/tex]

so i don't know how to solve it.
 
  • #60
how to get the last part?
 
  • #61
Look, transgalactic, this is screamingly obvious …

since f(0) = 0, what is [tex]\lim _{x->0}\frac{f(x) }{x}[/tex] the definition of? :frown:
 
  • #62
i don't know
i get 0 n the numerator and 0 in the denominator
i can see it in another way
[tex]
f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}
[/tex]
but i don't get a value
??
 
  • #63
transgalactic said:
i can see it in another way
[tex] f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}[/tex]

Yes, that's it!

Why do you have a mental block about these things?

As you say, that limit is f'(0) …

ok, now go back to posts #49-50 …
tiny-tim said:
ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ? :smile:
transgalactic said:
[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}
[/tex]

which = … ? :smile:
 
  • #64
[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}
[/tex]
but i was asked to calculate
and it doesn't give me a result
??
 
  • #65
tiny-tim said:
ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ? :smile:
transgalactic said:
[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}
[/tex]
but i was asked to calculate

Yes …
transgalactic said:
f(x) and g(x) are differentiable on 0
f(0)=g(0)=0

calculate
[tex]\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} [/tex]

which is the same as …
[tex]\lim _{x->0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}[/tex]

which is … ? :smile:
 
  • #66
i think
[tex]
\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}
[/tex]
but its not a result

??
 
  • #67
transgalactic said:
i think
[tex]
\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}
[/tex]
but its not a result

??

[tex]\frac{-f'(0)^2}{2!}+\frac{g'(0)^2}{2!}[/tex] actually


but why do you think that's not a result?
 
  • #68
because i was told
"calculate"
i here i have only an expression

??
 
  • #69
transgalactic said:
because i was told
"calculate"
i here i have only an expression

??

oh i see!

no, an expression is ok

i admit "calculate" usually means a number …

but it isn't an official word, it's just another way of saying "work out" :smile:

is that all that was bothering you?
 
  • #70
i haven't been given f'(0)

i was told that it was differentiable on point 0.
but i don't know what thing are given
so i can use them into the solution expression
??
 

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