When x approaches a, what is f, the limit of F is 0

When x approaches a, what is f, the limit of F is 0

Suppose f (a) = f (a) = 0 to make f (x) and f (x) continuous at point x = 0. Because Cauchy's mean value theorem requires two functions to be continuous in a closed interval, f (x) and f (x) may not be defined at point x = a, but LIM (x → a) f (x) = LIM (x → a) f (x) = 0. Therefore, in order to make f (x) and f (x) continuous at point x = a, f (a) = f (a) = 0 is defined

Let f '(0) = 1, then what is the limit of {f (0) - f (2)} / X when x approaches 0

When f (2) = f (0), the limit is 0
When f (2) ≠ f (0), the limit does not exist

Let f (x) be differentiable at x = a, and find the value of LIM x → a f (x) - f (a) / x-a

It's f '(a). Ah, the definition of derivative

Let f (x) be derivable at X., and find the value of LIM [h → 0] {f (X. - HX) - f (x)} / HX
Where HX = X-X.

Right? HX
HX = X-X., that is, x = X. + HX
After substitution
f（x.-hx）-f(x.+hx)/hx
[f（x.-hx）-f(x.)]+[f(x.)-f(x.+hx)]/hx
-{[f(x.)-f(x.-hx)]+[f(x.+hx)-f(x.)]}/hx
After grouping, it is obvious that if X. is differentiable, then the limit value - 2F '(X.)

Let y = f (x) be differentiable at x = 1 and f '(1) = 2, then Lim △ x → 0 [f (1 + 2 △ x) - f (1)] / △ X=
Seek the process~

Let f (x) = LIM (t → + ∞) (x / (2 + X Λ 2-e Λ (TX)), discuss the derivability of F (x)
Ask for detailed explanation

Let f (x) be differentiable in a neighborhood of x = 0, and lim - Intermediate derivative / x = 1, then does f (0) have an extremum? Lim tends to 0

Because the first derivative of LIM / x = 1
That is Lim f '(x) / x = 1
That is Lim [f '(x) - f' (0)] / (x-0) = 1
From the definition of derivative, we know that the derivative of F '(x) at x = 0 (i.e. the second derivative) f' '(0) = 1 > 0
So f '(x) is an increasing function near x = 0
So near x = 0,
When x > 0, f '(x) > F' (0) = 0, the function increases;
When x

If f (x) is differentiable in the neighborhood of x = x0, and f '(x0) = 0, LIM (x ~ x0) f' (x) = 1, can f (x) get the extremum at x = x0?

From the differential mean value theorem
f(x)-f(x0)=f'(ξ)(x-x0)
ξ belongs to (x0, x) (when x is less than x0, it is (x, x0))
Because LIM (x ~ x0) f '(x) = 1
For ε = 1 / 2, there exists δ > 0
|x-x0|0
So in (x0 - δ, x0 + δ)
All the points on the right side of F (x0) are larger than f (x0)
All the points on the left side of F (x0) are smaller than f (x0)
So it's not an extreme point

Let f (x) be differentiable at a and find the extremum Lim H-0 f (a) - f (A-H) / h

f '(a)

Let f (x) be differentiable, f '(0) = 0, LIM (x approaches 0) f' (x) / x = 2, then is f (0) its extremum, maximum or minimum?

LIM (x - > 0) f '(x) / x = 2 > 0, we can know from the sign preserving property of limit that in a δ neighborhood of x = 0, there must be a simple increase of x0, f (x)
If f '(0) = 0, we know that f (x) has a minimum at x = 0, that is, f (0) is its minimum