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

The problem of function discontinuity Let the function f (x) = [e ^ (1 / x) - 1] / [e ^ (1 / x) + 1], then x = 0 is the () A. Removable discontinuity B. jumping discontinuity C. infinite discontinuity D. oscillation discontinuity I wonder 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 numerator and denominator of F (x) are divided by e ^ 1 / X,
F (x) = [1-e ^ (- 1 / x)] / [1 + e ^ (- 1 / x)], f (x) → [1-0] / [1 + 0] = 1
Therefore, at x = 0, both the left limit and the right limit of F (x) exist, but they are not equal, which belongs to a jumping breakpoint

Function discontinuity problem! If there is a function y = Tan x, where - pi / 2 < x < pi / 2, are - pi / 2 and PI / 2 discontinuous points of function y? Why? Just after reading the reference book (page 62, Volume I, Tongji sixth edition of Advanced Mathematics), I found that x0 is the discontinuity of function y, provided that function y is defined in a de centered field of x0, so - pi / 2 and PI / 2 in the question are not the discontinuity of function y = Tan x (- pi / 2 < x < pi / 2)

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

A simple problem about function discontinuity If f (x) = (x-a of E) divided by [x (x-1)] has a removable discontinuity x = 1, find Changshu a, and ask if f (x) has any discontinuity, and if so, what type? (please write down the steps to explain)

f(X)＝（e^x－a)/[x（x－1)]
Because there is a removable discontinuity x = 1, when x tends to 1, the molecular limit = 0, so a = E
When x tends to 0, the denominator tends to 0, the molecule tends to 1-e, and the limit is infinity, so x = 0 is the infinite breakpoint

Function breakpoint problem set up f(x)=(e^1/x - 1)/(e^1/x + 1) Then x = 0 is what type of discontinuity of function f (x)? It's best to have process and explanation

f(x)=1-2/(e^1/x+1)
Notice that lime ^ 1 / x = + infinity when x approaches 0 on the right, and lime ^ 1 / x = 0 when x approaches 0 on the left
Therefore, 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
Therefore, x = 0 is the first type of breakpoint and jump breakpoint

Mathematical function discontinuity problem 【1】 Function y = 1 / in|x| How many discontinuities are there 【2】 Function y = [X-1] / [x ^ 2-2x-3] with {] discontinuity

【1】 Function y = 1 / in|x|
How many discontinuities are there
1, x = 0
【2】 Function y = [X-1] / [x ^ 2-2x-3] with {] discontinuity
The [] here is just ()
x^2-2x-3=0
x=3
x=-1
2 breakpoints

According to the definition of monotonicity of function, the function f is proved   (x) = - X3 + 1 is a subtractive function on (- ∞, + ∞)

Proof: proof 1: take x1, X2 and X1 < x2 in (- ∞, + ∞)
Then f (x2) - f (x1) = x13-x23 = (x1-x2) (X12 + x1x2 + X22)
∵x1＜x2，
∴x1-x2＜0．
When x1x2 ＜ 0, there is X12 + x1x2 + X22 = (x1 + x2) 2-x1x2 ＞ 0;
When x1x2 ≥ 0, X12 + x1x2 + X22 ＞ 0;
∴f（x2）-f（x1）=（x1-x2）（x12+x1x2+x22）＜0．
That is, f (x2) < f (x1)
Therefore, the function f (x) = - X3 + 1 is a subtractive function on (- ∞, + ∞)
Proof 2: take X1 and X2 in (- ∞, + ∞), and x1 ＜ X2,
Then f (x2) - f (x1) = x13-x23 = (x1-x2) (X12 + x1x2 + X22)
∵x1＜x2，
∴x1-x2＜0．
∵ X1 and X2 are not zero at the same time,
∴x12+x22＞0．
∵ X12 + X22 ＞ 1
2（x12+x22）≥|x1x2|≥-x1x2
∴x12+x1x2+x22＞0，
∴f（x2）-f（x1）=（x1-x2）（x12+x1x2+x22）＜0．
That is, f (x2) < f (x1)
Therefore, the function f (x) = - X3 + 1 is a subtractive function on (- ∞, + ∞)