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Advanced mathematical problems, limits of bivariate functions Find the limit of LIM (1 + XY) ^ (1 / (x + y)) x→0 y→0 The solution in the book is the original formula = Lim e ^ XY / (x + y) (x,y)→(0,0) How did the master get this step, I already know LIM (1 + x) ^ 1 / x = E x→0
For the limit of power exponential function, the logarithm is often taken, so it is transformed into e ^ [ln (1 + XY) / (x + y)]. With the equivalent infinitesimal ln (1 + x) ~ x (x → 0), ln (1 + XY) is replaced by XY, and
Original formula = Lim e ^ XY / (x + y)
For the function y = 3 times root x, it can be proved that it is continuous at (0,0) according to the existence of left and right limits and equal to f (0) How to prove that the root sign x of function y = 3 is not differentiable at (0,0)?
It is proved that the function y = f (x) = x ^ 1 / 3 is continuous in the interval (- ∞, + ∞), but it is not differentiable at point x = 0. Because there is [f (0 + H) - f (0)] / h = (H ^ (1 / 3) - 0) / h = 1 / h ^ (2 / 3) at point x = 0, the limit LIM (H → 0) [f (H + 0) - f (0)] / h = LIM (H → 0) 1 / h ^ (2 / 3) = + ∞, that is, the derivative
Given the function f (x) = (root sign (a ^ 2-x ^ 2)) / (absolute value (x + a) - a), then the set {A / F (x) is an odd function}=
F (x) = √ (a ^ 2-x ^ 2) / (|x + a | - a) in order to make the function meaningful, a ^ 2-x ^ 2 ≥ 0, that is, | x ≤| a | if a = 0, then x = 0, so | x + a | - A = 0, f (x) is meaningless, so a ≠ 0 because f (0) = | a | / (| a | - a) if A0, then f (x) / X is an odd function. Therefore, the set {a | f (x) is an odd function} = {
How to prove the continuity of partial derivative function of binary function
Generally, it is a piecewise function. For segments with continuous differentiable open intervals, its partial derivative can be obtained directly, and then its partial derivative value can be obtained by the definition method for the segment point or judged to be nonexistent. Therefore, whether the partial derivative is continuous at the segment point can be judged
Prove whether there is a function that satisfies: "it can be derived everywhere, but the derivative function is discontinuous everywhere" Because we already know that there is a function of "continuous everywhere, but non differentiable everywhere", but we can't find an argument on whether this function exists on the Internet
In fact, the continuous point set of the derivative of the differentiable function on the closed interval I must be dense on I! See the thinking question on page 55 of Zhou Minqiang's theory of functions of real variables. The general idea is as follows: first, remember F_ N (x) = n [f (x + 1 / N) - f (x)], then f_ N is a continuous function. Since f is everywhere differentiable, for every x ∈ I
Let the function f (x) be differentiable at x = A and f '(a) not equal to 0. Find the limit of the 1 / x power of [f (a + x) / F (a)] when x tends to 0
X → 0lim [f (a + x) / F (a)] ^ (1 / x) = Lim e ^ ln [f (a + x) / F (a)] ^ (1 / x) = e ^ Lim ln [f (a + x) / F (a)] ^ (1 / x) consider Lim ln [f (a + x) / F (a)] ^ (1 / x) = Lim [LNF (a + x) - LNF (a)] / X according to the definition of derivative = [LNF (x)] '|x = a = f' (a) / F (a). Therefore, the original limit = e ^ [f '(a
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