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

[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

<br /> f&#039;(x)=\lim _{x-&gt;0}\frac{f(x)-f(0)}{x-0}=\lim _{x-&gt;0}\frac{f(x)-0}{x-0}<br /> <br />
this is the definition of the derivative
i don't know how to continue

you said also "Difference quotient" so i used
<br /> f&#039;(x)=\lim _{h-&gt;0}\frac{f(x+h)-f(x)}{h}<br />
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&#039;(x)=\lim _{x-&gt;0}\frac{f(x)-f(0)}{x-0}=\lim _{x-&gt;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&#039;(0)=\lim _{x-&gt;0}\frac{f(x)-f(0)}{x-0}=\lim _{x-&gt;0}\frac{f(x)-0}{x-0}=\lim _{x-&gt;0}\frac{f(x)}{x}

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

\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{-1}{2}\,\left(\lim _{x-&gt;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
<br /> \ \frac{-1}{2}\,\left(\lim _{x-&gt;0}\frac{f(x) }{x}\right)^2=\ \frac{-1}{2}\,\left(\lim _{x-&gt;0}\frac{0 }{0}\right)^2=<br />

so i don't know how to solve it.

 

[transgalactic]

how to get the last part?

 

[tiny-tim]

Look, transgalactic, this is screamingly obvious …

since f(0) = 0, what is \lim _{x-&gt;0}\frac{f(x) }{x} the definition of?

 

[transgalactic]

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
??

 

[tiny-tim]

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 …

ok, so what can you say about \lim _{x-&gt;0}\frac{cos(f(x)) - 1}{x^2} ? ​

<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 = … ?

 

[transgalactic]

<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
??

 

[tiny-tim]

ok, so what can you say about \lim _{x-&gt;0}\frac{cos(f(x)) - 1}{x^2} ? ​

<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 …

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 … ?

 

[transgalactic]

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

??

 

[tiny-tim]

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?

 

[transgalactic]

because i was told
"calculate"
i here i have only an expression

??

 

[tiny-tim]

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"

is that all that was bothering you?

 

[transgalactic]

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
??

 

[tiny-tim]

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! ​

 

[transgalactic]

thanks:)