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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 said: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]
transgalactic said: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]
??
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]
transgalactic said:the expansion for cos x around 0 is
[tex]cosx=1+0-\frac{x^2}{2!}[/tex]
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 .
transgalactic said: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.
[tex]\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}[/tex]
i gave every option i can think of.D H said:So what is f(x) near zero?
Tiny-tim, please do not give this away.
transgalactic said:i don't 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 don't know how to continue
[tex]=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2[/tex]transgalactic said:[tex]\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}[/tex]
transgalactic said:i can see it in another way
[tex] f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}[/tex]
tiny-tim said:ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ?
transgalactic said:[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}
[/tex]
tiny-tim said:ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ?
transgalactic said:[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
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]
transgalactic said:i think
[tex]
\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}
[/tex]
but its not a result
??
transgalactic said:because i was told
"calculate"
i here i have only an expression
??