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

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Homework Help Overview

The discussion revolves around calculating the limit of the expression \(\lim _{x \to 0} \frac{\cos f(x) - \cos g(x)}{x^2}\), where \(f(x)\) and \(g(x)\) are differentiable at \(0\) and both equal to \(0\) at that point. Participants explore the implications of Cauchy's Mean Value Theorem and the conditions under which L'Hôpital's rule can be applied.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the applicability of L'Hôpital's rule and question whether \(f\) and \(g\) are twice differentiable. There are attempts to express the limit in terms of derivatives and to clarify the definitions of differentiability and continuity at \(0\).

Discussion Status

The conversation is ongoing, with various interpretations of the problem being explored. Some participants have provided hints and suggestions for alternative approaches, such as using series expansions, while others emphasize the need for clarity regarding the differentiability of \(f\) and \(g\) around \(0\).

Contextual Notes

There is a notable emphasis on the technicalities of differentiability, particularly that \(f\) and \(g\) are not necessarily differentiable around \(0\), which complicates the application of certain mathematical tools like L'Hôpital's rule.

  • #61
Look, transgalactic, this is screamingly obvious …

since f(0) = 0, what is \lim _{x->0}\frac{f(x) }{x} the definition of? :frown:
 
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  • #62
i don't know
i get 0 n the numerator and 0 in the denominator
i can see it in another way
<br /> f&#039;(0)=\lim _{x-&gt;0}\frac{f(x)-f(0) }{x-0}<br />
but i don't get a value
??
 
  • #63
transgalactic said:
i can see it in another way
f&#039;(0)=\lim _{x-&gt;0}\frac{f(x)-f(0) }{x-0}

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 \lim _{x-&gt;0}\frac{cos(f(x)) - 1}{x^2} ? :smile:
transgalactic said:
<br /> \lim _{x-&gt;0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x-&gt;0}\frac{\ -\ (f(x))^2 }{x^22!}<br />

which = … ? :smile:
 
  • #64
<br /> \lim _{x-&gt;0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x-&gt;0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f&#039;(0)^2}{2!}<br />
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 \lim _{x-&gt;0}\frac{cos(f(x)) - 1}{x^2} ? :smile:
transgalactic said:
<br /> \lim _{x-&gt;0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x-&gt;0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f&#039;(0)^2}{2!}<br />
but i was asked to calculate

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

calculate
\lim _{x-&gt;0}\frac{cosf(x)-cosg(x)}{x^2}

which is the same as …
\lim _{x-&gt;0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}

which is … ? :smile:
 
  • #66
i think
<br /> \frac{-f&#039;(0)^2}{2!}+\frac{-g&#039;(0)^2}{2!}<br />
but its not a result

??
 
  • #67
transgalactic said:
i think
<br /> \frac{-f&#039;(0)^2}{2!}+\frac{-g&#039;(0)^2}{2!}<br />
but its not a result

??

\frac{-f&#039;(0)^2}{2!}+\frac{g&#039;(0)^2}{2!} 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
??
 
  • #71
transgalactic said:
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
??

yes yes yes!

no problemo!

go for it! :smile:
 
  • #72
thanks:)
 

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