The problem of discontinuous point of function Let f (x) = [e ^ (1 / x) - 1] / [e ^ (1 / x) + 1], then x = 0 is () A. Removable discontinuity B. jumping discontinuity C. infinite discontinuity D. oscillation discontinuity I want to know why

The problem of discontinuous point of function Let f (x) = [e ^ (1 / x) - 1] / [e ^ (1 / x) + 1], then x = 0 is () A. Removable discontinuity B. jumping discontinuity C. infinite discontinuity D. oscillation discontinuity I want to know why

f(x)=[e^(1/x)-1]/[e^(1/x)+1],
When x → 0 -, 1 / X → - ∞, e ^ 1 / X → 0, f (x) → [0-1] / [0 + 1] = - 1
When x → 0 +, 1 / X → + ∞, e ^ 1 / X → + ∞, e ^ (- 1 / x) → 0, the denominator of F (x) is divided by e ^ 1 / X,
It is reduced to f (x) = [1-e ^ (- 1 / x)] / [1 + e ^ (- 1 / x)], f (x) → [1-0] / [1 + 0] = 1
So at x = 0, the left limit and the right limit of F (x) exist, but they are not equal

Function discontinuity point problem!
If there is a function y = Tan x, where - pi / 2 < x < pi / 2, are - pi / 2 and PI / 2 discontinuities of function y? Why?
Just after reading the reference book (Tongji sixth edition, Vol. 1, page 62), I found that x0 is the discontinuity of function y, provided that function y is defined in a certain field of x0, so - pi / 2 and PI / 2 in the question are not the discontinuities of function y = Tan x (- pi / 2 & lt; X & lt; PI / 2)

Of course, it is not a discontinuity point. The first thing a function needs to determine is the domain, which does not contain ± pi / 2

The problem of function discontinuity
set up
f(x)=(e^1/x - 1)/(e^1/x + 1)
Then x = 0 is the type of discontinuity of F (x)?
It's better to have process and explanation

f(x)=1-2/(e^1/x+1)
Note that when the right side of x approaches 0, lime ^ 1 / x = + infinity, and when x approaches 0 from the left side, lime ^ 1 / x = 0
So when x tends to 0 from the right, f (x) tends to 1-0 = 1, and when x tends to 0 from the left, f (x) tends to 1-2 = - 1
So x = 0 is the first kind of discontinuity, and also the jump discontinuity

Break point problem of mathematical function
【1】 Function y = 1 / in|x|
How many discontinuities are there
【2】 The function y = [X-1] / [x ^ 2-2x-3] has {]

【1】 Function y = 1 / in|x|
How many discontinuities are there
1, x = 0
【2】 The function y = [X-1] / [x ^ 2-2x-3] has {]
It's just () here
x^2-2x-3=0
x=3
x=-1
2 discontinuities

How to find the discontinuous point of function

1. General artificial function, more piecewise function, abstract function, such discontinuities are expressed in the topic, very easy to find
2. The discontinuity point of natural function generally starts from the domain of definition, that is, where is the so-called function expression meaningless? Such a point is the discontinuity point

The first kind of discontinuity of derivative

If x0 is the discontinuous point of function f (x), but both the left limit and the right limit exist, then x0 is called the first kind of discontinuity of function f (x). Let y = f (x) be defined in a neighborhood of point x0, if the limit of function f (x) exists when x → x0 and is equal to its function value f (x0) at point x0, that is, LIM (x → x0)

What are the meanings of the first type of discontinuity and the second type of discontinuity

The first type is the one with both left and right limits

What is the first kind of discontinuity in definite integral?
The existence theorem of definite integral: if f (x) is continuous in [a, b], or there are at most finite discontinuities of the first kind, then f (x) is integrable in [a, b]
What is the "first kind of discontinuity" here?

Jumping discontinuities are the first kind of discontinuities
If the left and right limits of a function are equal, but the value of the function is meaningless, it is called a removable breakpoint, and if the function is not equal, it is called a jumping breakpoint

Removable discontinuities in the first kind of discontinuities in Mathematics
Why is it necessary to supplement or modify the definition only for the removable discontinuities

Because of people's needs, it's defined as that

If the limit of a function exists at one point, but it is discontinuous at this point, then the point is the first kind of discontinuity
Right or wrong?

Wrong!
If x0 is the discontinuous point of function f (x), but both the left limit and the right limit exist but are not equal, then x0 is called the first kind of discontinuity point of function f (x)