F (x) = TaNx, there is an infinite breakpoint in its definition domain from negative infinity to positive infinity. Why do you say that "all elementary functions are continuous in their definition interval"?

F (x) = TaNx, there is an infinite breakpoint in its definition domain from negative infinity to positive infinity. Why do you say that "all elementary functions are continuous in their definition interval"?

To raise this question shows that you are thinking, but you should pay attention to the key description in the topic. All elementary functions are continuous in their "definition interval", f (x) = TaNx, and there are infinite breakpoints in their "definition domain" from negative infinity to positive infinity. About the difference between the two: definition interval: it is only a range, which represents the function

Tongji high number: oscillation breakpoint, why "at least one of the left and right limits does not exist"? Tongji advanced mathematics books: y = sin (1 / x), x = 0 is called the oscillation breakpoint My understanding: "X --" 0, sin (1 / x) - "0", the limit of the function is 0, that is, the function is infinitesimal. Why does the textbook say that it is "at least one of the left and right limits does not exist"? Where is my misunderstanding?

X -- (1 / x) of 0 - ∞ sin θ You can draw a graph, sin at infinity θ It is not a fixed value, but oscillates between [- 1,1], and the value is not unique

Breakpoints can be removed for high number limit problems Is the discontinuity of the function y = sinxsin (1 / x) 0? Is it OK to go? What about 1 / x? Isn't 1 / 0 meaningless?

It's removable, because when x goes to zero, the limit of Y is 0

The right limit is infinite, and the left limit is the break point of oscillation change

At least one of the left and right limits does not exist, so it is the second type of discontinuity, in which the limit is infinite. As you see here, it is called infinite discontinuity. Add: you can't call it oscillatory discontinuity here, because the definition of oscillatory discontinuity is: "the function has no definition at this point. When the independent variable approaches this point, the value of the function is in two constants

Is there a limit value at the oscillation discontinuity It is mentioned in the book that sin (1 / x) and COS (1 / x) are oscillatory discontinuities at x = 0. Although these functions are discontinuous at x = 0, does there exist a limit value at x = 0? Does the limit value of X * cos (1 / x) exist (when X - > 0)?

Sin (1 / x) has no limit at x = 0. You think, when it approaches 0, 1 / X tends to positive infinity. Similarly, on the coordinate axis, when 1 / X in sin (1 / x) tends to positive infinity, does sin (1 / x) always fluctuate between 1 and - 1 and can't stop? Of course, cos (1 / x) is also a truth
As for X * cos (1 / x), its limit is 0 at 0. The reason is that although cos (1 / x) has no limit, it is bounded. The product of a bounded function and an infinitesimal must be an infinitesimal value. Here, it is 0. (0 is also infinitesimal)

What are differentiable and non differentiable functions? What are the conditions?

Let y = f (x) be a univariate function. If y has a derivative y '= f' (x) at x = x [0], y is said to be differentiable at x = x [0]
Conditions: 1) if f (x) is continuous at x0, when a tends to 0, [f (x + a) - f (x)] / A has a limit, then f (x) is said to be differentiable at x0
(2) If f (m) is differentiable for any point m on interval (a, b), then f (x) is said to be differentiable on (a, b)
Discontinuous functions must be non differentiable
Another is that although the function is continuous, the left derivative and right derivative at a certain point are not equal. For the problems of left derivative and right derivative, please refer to the mathematical analysis of the University