X = 0 is the () breakpoint of function f (x) = 1 / 1-2 ^ X

X = 0 is the () breakpoint of function f (x) = 1 / 1-2 ^ X

I'm glad to answer for you
The answer is: infinite discontinuity
Because the left and right limits of F (x) do not exist
If you still don't understand, welcome to ask~

Find the discontinuity of function f (x) = (1 + x) ^ [x / Tan (x - π)] in (0,2 π) And judge its type

π / 2,3 π / 2 are the first kind of removable breakpoints (limits exist and are all 1)
π is the second type of infinite discontinuity (x approaches infinity from positive direction and 0 from negative direction)

The continuous interval of function f (x) = (X-6) / (x ^ 2-4x-12) is __, What is the discontinuity?

f(x)=(x-6)/(x^2-4x-12)=(x-6)/(x-6)(x+2)
The continuous interval of function f (x) = (X-6) / (x ^ 2-4x-12) is (- infinite, - 2) ∪ (- 2, + infinite) and the discontinuity point is x = - 2

If y = f (x) has zero on [- 1,1], find the value range of real number a Given the function f (x) = x ^ 2-4x + A + 3, G (x) = MX + 5-2m, when a = 0, for any x1 ∈ [1,4], there is always x2 ∈ [1,4], so that f (x1) = g (x2) holds, and find the range of M If the value range of function y = f (x) (x ∈ [T, 4]) is interval D, is there a constant t so that the length of interval D is 7-2t? If so, find the value of T, and explain why it does not exist

It is calculated that f (2) = A-1, f (1) = a, f (4) = a + 3
g(1)=5-m g(4)=5+2m
Therefore, f (x) ∈ [A-1, a + 3]
m>0,g(x)∈[5+2m,5-m]
m>0,g(x)∈[5-m,5+2m]
5-m3 6t>=0,7-2t>3,D=3
Therefore, it does not exist
When t

Let f (x) = (x ^ 2-1) / (x ^ 3-3x + 2), point out the discontinuities of the function, and explain what kind of discontinuities these discontinuities belong to There's something wrong with it F (x) = f (x) = (x ^ 2-1) / (x ^ 2-3x + 2), sorry

f(x)=(x^2-1)/(x^2-3x+2)
=(x+1)(x-1)/[(x-2)(x-1)]
=(x+1)/(x-2)
=1 + 3 / (X-2) (x ≠ 1 and ≠ 2)
So the discontinuities are x = 1 and x = 2
Are the second type of breakpoints

The first type of discontinuity of function f (x) = x / SiNx on R is?

All discontinuities of the function are {x|x = k π, K ∈ Z}. When x → 0, X / SiNx → 1, so x = 0 is the first type of discontinuity of X / SiNx. At other discontinuities, the function has no limit, so all discontinuities except x = 0 are the second type of discontinuity of the function