differantiation question ..

tiny-tim

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

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

[tex]\frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2[/tex]

which = … ?

transgalactic

i dont know whats the value of the limit

f(x) goes goes to f(0) but not equals f(0) so i dont know whats 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 dont know what is the value of f'(0)
i know that f(0)=0

[tex]
f'(x)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}

[/tex]
this is the definition of the derivative
i dont know how to continue

you said also "Difference quotient" so i used
[tex]
f'(x)=\lim _{h->0}\frac{f(x+h)-f(x)}{h}
[/tex]
but i dont have any values for it.

transgalactic

i gave every option i can think of.
i dont know.

tiny-tim

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):

[tex]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}[/tex]

ok, so now you have …
[tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}\ =[/tex]
[tex]=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2[/tex]

= … ?

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
[tex]
\ \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=
[/tex]

so i dont 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 [tex]\lim _{x->0}\frac{f(x) }{x}[/tex] the definition of?

transgalactic

i dont know
i get 0 n the numerator and 0 in the denominator
i can see it in another way
[tex]
f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}
[/tex]
but i dont get a value
??

tiny-tim

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 …
which = … ?

transgalactic

[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}
[/tex]
but i was asked to calculate
and it doesnt give me a result
??

tiny-tim

Yes …
which is the same as …
[tex]\lim _{x->0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}[/tex]

which is … ?

transgalactic

i think
[tex]
\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}
[/tex]
but its not a result

??

tiny-tim

[tex]\frac{-f'(0)^2}{2!}+\frac{g'(0)^2}{2!}[/tex] actually
…

but why do you think that's not a result?

transgalactic

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

??