Let f (x) be continuous on [0, ∞), and when x > 0, 0

Just prove that f (x) is incrementally bounded:
in fact,
1) F (x) is obtained when the derivative is greater than 0;
2) F (x) has an upper bound:
Use f (x) = f '(s) to integrate from 1 to x, plus f (1)
Because f '(x)

Does the left limit of F (x) = 2 ^ (1 / x) exist at x = 0, or does the right limit exist?

Left limit exists, 0

How to prove that the limit of x ^ 2 when x tends to 2 is 4? It is proved by the strict definition of limit Especially that idea, how do you think of taking δ For that value

It can be proved by the limit of the function

Find the left and right limits when f (x) = x / x, H (x) = |x| / X. and explain whether their limits exist when x tends to 0? =

F (x) = x / x, left limit = right limit = 1
H (x) = |x| / x, left limit = - 1, right limit = 1, limit does not exist

If the function f (x) has a limit at some point, it can be derived at that point

Of course not. Whether the limit exists at a certain point means whether it is continuous. If the left and right limits exist and are equal and equal to the function value at that point, the function is continuous. But if the derivative exists, the function must be continuous, so we can know the existence of the limit of the function

Given that ABCD is in an equal ratio sequence, and the vertex of curve y = x ^ 2-2x + 3 is (B, c), then what is ad equal to? This is an optional question, A.3 B.2 C.1 D. - 2 This is the Hainan college entrance examination question. There is no 4 you calculated

The upstairs is wrong
Vertex should be (- B / 2a, (4ac-b ^ 2) / 4A)
So the vertex coordinates are (1,2)
Ad = 2 because ad = BC

Let f (x) be continuous on [0, ∞), and when x > 0, 0