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
  • #1
transgalactic
1,395
0
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]
??

it looks like cauchys mean value theorem
F(x)=cos(f(x))
G(x)=cos(g(x))
what to do next??
 
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  • #2
Welcome to the second millenium!

Hi transgalactic! Congrats on your 1001st post! :smile:
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]

Hint: use L'Hôpital's rule … twice! :wink:
 
  • #3
tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.
 
  • #4
always look on the bright side of life … ♩ ♫ ♪ ♫ ♬ ♩

HallsofIvy said:
tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.

oh tush, mr glass-half-empty …

:biggrin: we don't know that they aren't! :biggrin:

:smile: happy days are here again
the skies above are clear again
let us sing a song of cheer again
happy days are here again! :smile:
 
  • #5
[tex]
\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}
[/tex]
so the solution is 0
??
 
  • #6
transgalactic said:
[tex]
\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}
[/tex]
so the solution is 0
??

uhh? :confused:

however did you get that?

always show your full calculations! :rolleyes:
 
  • #7
[tex]
\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+f'(x)cos(f(x))-f''(x)sing(x)+f'(x)cosg(x)}{2}
[/tex]
??
 
  • #8
That's better! :smile:

And if f(0)=g(0)=0, then that = … ?
 
  • #9
Transgalactic, as much as I hate to pour cold water on the party, you started by saying "if f and g are differentiable at 0". Is that all you are given or are you given that the second derivative exists?
 
  • #10
its said that" they are differentiable on 0

but not necessarily differentiable around 0
 
  • #11
HallsofIvy said:
Transgalactic, as much as I hate to pour cold water on the party, you started by saying "if f and g are differentiable at 0". Is that all you are given or are you given that the second derivative exists?
transgalactic said:
its said that" they are differentiable on 0

but not necessarily differentiable around 0

This is a bit of a technicality … we'll discuss why in a moment …

but first, just answer my previous question: what is that limit equal to at x = 0?

(oh, btw, just noticed :redface: … you should have had an f'(x)2 somewhere in that :wink:)
 
  • #12
i fixed it to
[tex]
\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+(f'(x))^2cos(f(x))-f''(x)sing(x)+(f'(x))^2cosg(x)}{2}
[/tex]

after i did lhopital twice i got
[tex]
\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-f''(x)sin0+(f'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(f'(x))^2}{2}
[/tex]

??
 
  • #13
HallsofIvy said:
Transgalactic, as much as I hate to pour cold water on the party, you started by saying "if f and g are differentiable at 0". Is that all you are given or are you given that the second derivative exists?

I'm with Halls. You only know the derivatives exist AT zero. You can't even apply l'Hopital the first time, you can't take the limit of something you only know exists at a single point. I'd suggest using the series expansion for cos(x) and difference quotients.
 
  • #14
what is the link between the ability of doing lhopital law
and differentiability of a function?
 
  • #15
Take f(x)=x^2*sin(1/x^2), f(x)=0. That's differentiable at 0. In fact, it's differentiable everywhere. But it's not continuously differentiable at 0. You can't do anything with l'Hopital at 0. But that doesn't mean your limit doesn't exist. So the problem with l'Hopital is more than 'technical'.
 
  • #16
transgalactic said:
i fixed it to
[tex]
\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+(f'(x))^2cos(f(x))-f''(x)sing(x)+(f'(x))^2cosg(x)}{2}
[/tex]

after i did lhopital twice i got
[tex]
\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-f''(x)sin0+(f'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(f'(x))^2}{2}
[/tex]

??

oh, transgalactic, however do you manage to make these mistakes?

you started with an expression that was half f and half g,

but you ended losing all the g :cry:

try again :smile:

(and then we'll sort out the technicalitites :wink:)
 
  • #17
fixed it
[tex]
\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(g'(x))^2}{2}
[/tex]

??
 
Last edited:
  • #18
transgalactic said:
fixed it
[tex]
\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(g'(x))^2}{2}
[/tex]

??

much better! :smile:

(though you still need to fiddle around with those pluses and minuses :wink:)

EDIT: oh, and since f'(0) and g'(0) are both defined as existing, is there any reason why you can't use them at the end? :smile:
 
Last edited:
  • #19
fixed it
[tex]
\lim _{x->0}\frac{-f''(x)sin(0)-(f'(x))^2cos(0)+g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{-(f'(x))^2+(g'(x))^2}{2}
[/tex]

??
 
Last edited:
  • #20
what now??
 
  • #21
Transgalactic, could you state the question exactly as written in your text? In particular, the phrase "f(x) and g(x) are differentiable on 0" does not make much sense (to me, at least). Does the text say something slightly different than "on 0"?
 
  • #22
f(x) and g(x) are differentiable on 0.
f(0)=g(0)=0
calculate "the presented limit"
notice:
f(x) and g(x) are not necessarily differentiable around 0
 
  • #23
transgalactic said:
fixed it
[tex]
\lim _{x->0}\frac{-f''(x)sin(0)-(f'(x))^2cos(0)+g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{-(f'(x))^2+(g'(x))^2}{2}
[/tex]

??

yes! :smile:

and now how about using f'(0) and g'(0)?​
 
  • #24
how??
its differentiable on
but i don't know the final value of the derivative around
??
 
  • #25
transgalactic said:
how??
its differentiable on
but i don't know the final value of the derivative around
??
What you have written assumes the derivative exists around 0. (In fact, as I have pointed out, this whole approach assumes f and g are twice differentiable in some neighborhood of 0)

If the derivative does exist in some neighborhood of 0, then, although a derivative is not necessarily continuous, it does satisfy the "intermediate value property" so [itex]\lim_{x\rightarrow 0} f'(x)= f'(0)[/itex] and [itex]\lim_{x\rightarrow 0} g'(x)= g'(0)[/itex]. I presume you do not know the actual values of f'(0) and g'(0) but at least
[tex]\frac{g'(0)^2- f'(0)^2}{2}[/tex]
is simpler than what you have.
 
Last edited by a moderator:
  • #26
transgalactic said:
f(x) and g(x) are differentiable on 0.
f(0)=g(0)=0
notice:
f(x) and g(x) are not necessarily differentiable around 0
So, what does "f(x) and g(x) are differentiable on 0" mean to you? Say so in mathematics. Understanding what this means is key to answering this question.
 
  • #27
D H said:
So, what does "f(x) and g(x) are differentiable on 0" mean to you? Say so in mathematics. Understanding what this means is key to answering this question.

that means that by definition of a derivative the limit of g(x) and f(x) gives me a finite solution.
??
 
  • #28
Say it in math, not words.
 
  • #29
[tex]
\lim_{h->0}\frac{f(x+h)-f(x)}{h}=const\\
[/tex]
[tex]
\lim_{h->0}\frac{g(x+h)-g(x)}{h}=const
[/tex]

??
 
  • #30
No. The problem definitely does not say that f(x) and g(x) are straight lines, which is exactly what you have written in this last post.
 
  • #31
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]
 
  • #32
It's not a constant. Stop doing that.
 
  • #33
ok its not a constant i have written my words into math
what is the next step??
 
  • #34
What does the fact that the derivatives exist at x=0 tell you about the value of the functions for x near zero?
 
  • #35
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]
??
 

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