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

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SUMMARY

The forum discussion focuses on calculating the limit \(\lim _{x \to 0} \frac{\cos f(x) - \cos g(x)}{x^2}\) using Cauchy's Mean Value Theorem. Participants suggest using L'Hôpital's rule, but caution that the differentiability of functions \(f\) and \(g\) at points other than zero is not guaranteed. The final expression simplifies to \(\frac{(g'(0))^2 - (f'(0))^2}{2}\), assuming the derivatives exist and are defined at zero.

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  • #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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