f 0 exists, and lim x tends to 0 1 x f x f x 3 a, find f 0

The limit exists, and the denominator = x tends to 0, so the numerator must also tend to 0, so it can be obtained using Robida's rule.

Molecule d(f(x) -f(x 3)) dx = f'(x) -f'(x/3)/3

Denominator dx dx = 1

So take the limit to get f'(0) -f'(0 3) 3 = a, so f'(0) -f'(0/3)/3 = f'(0) *2 3 = a so f'(0) = 3a/2

2l is wrong.

The title only says f'(0) exists, and does not say f'(x) exists, and the derivative of f(x) cannot be directly obtained.

However, if it is a multiple-choice question that does not require a process, the method of pressing 2L is relatively simple.

Derivatives you're enough!

Here come and watch silently... t t

This is not a simple question.

You see that the High Magician will not ...

This is the question of whether the derivative function is necessarily continuous.

That statement is indeed wrong.

f(x) is derivable at x0.

It is not possible to derive f'(x) Continuous at x0.

Example functions. f(x)=(x^2)

sin(1/x)

x is not equal to 0, f(0) = 0.

The landlord can consider this function.

This function is continuously derivable.

But you'll find out.

f'(0)=0

But lim [x 0] f'(x) does not exist.

f'(x) at x=0.

Discontinuous.

When x tends to 0, lim f(x) x=1 is derived from the Lopida's rule, which is derived from the numerator and denominator at the same time.

When x tends to 0, lim f(x) x=1=f'(x) 1 so f'(0)=1, so that f(x)=f(x) -x is clearly f(0)=0

Get f'(x)=f '(x) -1

So f'(0)=f '(0) -1=0, and f''x) >0, i.e. f'(x) monotonically increasing, and f'(0)=1, so when x>0, f'x) >0, i.e. f'(x)=f '(x) -1>0, so f(x) increases monotonically when greater than 0. x

lim(x->0)[f(x)-f(x/3)]/xlim(x->0)[f(x)-f(0)+f(0)-f(x/3)]/xlim(x->0)[f(x)-f(0)]/x-0)-lim(x->0)[f(x/3)-f(0)]/x

f'Split(0)-lim(x-> dongyuantong0)(1 Natan3)*[f(x 3)-f(0)] x 3-0).

f'(0)-1/3*lim(x/3->0)[f(x/3)-f(0)]/x/3-0)

f'(0)-1/3*f'(0)

2/3*f'(0)

i.e. 2 3*f'(0)=a

So f'(0)=3a/2

It should be lim( x tends to 0) f (x0 + 2 x) x = lim ( x tends to 0) f (x0 + 2 x) -f (x0) contains noisy + f (x0) x = 2 lim ( x tends to 0) f (x0 + 2 x) -f (x0) 2 x + f (x0) x = 2 f'(x0)+f'(x0)=12

Then find lim( x tends to 0)f(x0) x=f'(x0) is because the upper and lower levels of the talk are infinitesimal quantities, and the Robbie curl can be used to reach the rule.

I think so.

Utilize the definition of the derivative f'(x0)=lim [f(x)-f(x0)]/x-x0) .The limit process is x x0, so lim[ f(x0-x)-f(x0)] xLet t=x0-x, when x 0 there is t x0=lim [f(t)-f(x0)] x0-t]=-lim [f(t)-f(x0)] t-x0].

The limit process is t x0 = -f'(x0)..

The title is lim[f(x)-f(-x)] x exists, right?For example: f(x)=x+1, then f(-x)=-x+1

lim[f(x)-f(-x)] x=lim2x x=2, the limit exists. And there is no f(0)=0.

I'm afraid you're ignoring the other conditions.

If the problem is lim f(x)-[f(-x) x] exists, then it is very easy to do. Left = f(0)-lim[f(-x) x] exists, and it is easy to get limf(-x)=f(0)=0

f(x) and x are both 0 at x=0, so the above limits can be solved using the Robi Tower rule.

lim[f(x)/x]=lim[f'(x)/1]=f'(0)=0, in the above formula, x 0 should be added to the first two steps