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

[tiny-tim]

Look, transgalactic, this is screamingly obvious …

since f(0) = 0, what is $$\lim _{x->0}\frac{f(x) }{x}$$ 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
$$f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}$$
but i don't get a value
??

 

[tiny-tim]

i can see it in another way
$$f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}$$

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

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

which = … ?

 

[transgalactic]

$$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}$$
but i was asked to calculate
and it doesn't give me a result
??

 

[tiny-tim]

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

$$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}$$
but i was asked to calculate

Yes …

f(x) and g(x) are differentiable on 0
f(0)=g(0)=0

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

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

which is … ?

 

[transgalactic]

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

??

 

[tiny-tim]

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

??

$$\frac{-f'(0)^2}{2!}+\frac{g'(0)^2}{2!}$$ 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:)