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

[transgalactic]

is that ok??
$$\lim_{x->0}\frac{f(x)-f(0)}{x}=const\\$$
$$\lim_{x->0}\frac{g(x)-g(0)}{x}=const$$

 

[D H]

It's not a constant. Stop doing that.

 

[transgalactic]

ok its not a constant i have written my words into math
what is the next step??

 

[D H]

What does the fact that the derivatives exist at x=0 tell you about the value of the functions for x near zero?

 

[transgalactic]

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
$$\lim_{x->0^+}\frac{f(x)-f(0)}{x}=\lim_{x->0^-}\frac{f(x)-f(0)}{x}=f(0)=0$$
??

 

[D H]

You aren't telling me anything you didn't say in post 29, and that last bit, that the derivatives are zero, is wrong.

 

[transgalactic]

sorry i ment to write the definition of continuity
$$\lim_{x->0^+}f(x)=\lim_{x->0^-}f(x)=f(0)=0$$??

 

[D H]

Stop with the "??" stuff, please. And please answer the question raised in post #34.

 

[transgalactic]

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.

 

[D H]

Say that mathematically.

 

[transgalactic]

$$\lim_{x->0^+}f(x)=\lim_{x->0^-}f(x)=f(0)=0<br />$$

 

[D H]

What is the value of f(x) for x near zero?

 

[transgalactic]

the value of f(x) near zero is close to 0
i can't imagine anything else

 

[D H]

What does $$f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x}$$ tell you about f(x) when x is close to but not equal to zero?

 

[transgalactic]

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)

 

[D H]

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?

 

[tiny-tim]

is that ok??
$$\lim_{x->0}\frac{f(x)-f(0)}{x}=const\\$$
$$\lim_{x->0}\frac{g(x)-g(0)}{x}=const$$

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
$$\lim_{x->0^+}\frac{f(x)-f(0)}{x}=\lim_{x->0^-}\frac{f(x)-f(0)}{x}=f(0)=0$$
??

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 $$\lim_{x->0^+}\frac{f(x)-f(0)}{x}=\lim_{x->0^-}\frac{f(x)-f(0)}{x}$$ = f'(0) = 0

(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? )

hmm … where had we got to?

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
$$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2}$$

and let's rewrite that as

$$\lim _{x->0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}$$

(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 …

$$\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}$$

what expansion or approximation do you know for cos(f(x)) - 1 when x -> 0? ​

 

[transgalactic]

the expansion for cos x around 0 is
$$cosx=1+0-\frac{x^2}{2!}$$
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.

 

[tiny-tim]

the expansion for cos x around 0 is
$$cosx=1+0-\frac{x^2}{2!}$$
but we need to substitute f(x) instead of x in cos x …

(why did you write +0?

did you mean O(x⁴)?)​

Yes … you do need to substitute f(x) … so it's
$$cos(f(x))=1\ -\ \frac{(f(x))^2}{2!}$$

… 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 $$\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}$$ ? ​

 

[transgalactic]

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.
$$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}$$

 

[D H]

So what is f(x) near zero?

Tiny-tim, please do not give this away.

 

[tiny-tim]

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.
$$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}$$

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

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

$$\frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2$$

which = … ?

 

[transgalactic]

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
?

 

[Dick]

Difference quotient for f'(0).

 

[tiny-tim]

I think what Dick is getting at is, what is the definition of f'(0)?

 

[transgalactic]

i don't know what is the value of f'(0)
i know that f(0)=0

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

you said also "Difference quotient" so i used
$$f'(x)=\lim _{h->0}\frac{f(x+h)-f(x)}{h}$$
but i don't have any values for it.

 

[transgalactic]

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.

 

[tiny-tim]

i don't know what is the value of f'(0)
i know that f(0)=0

$$f'(x)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}$$
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):

$$f'(0)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}=\lim _{x->0}\frac{f(x)}{x}$$

ok, so now you have …
$$\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}\ =$$

$$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}$$

$$=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2$$

= … ?

 

[transgalactic]

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
$$\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{0 }{0}\right)^2=$$

so i don't know how to solve it.

 

[transgalactic]

how to get the last part?