Advanced Mathematics questions about discontinuities and continuities

1.(x) and f(x) are defined in the domains (- f( (x)) may be continuous.

For example, (1) f(x)=| x |,x)= x+1 (x>0) ,x)= x-1 (x<0) ,0)=1 ,x=0 is the hop break point of (x), f( (0)) =1, x tends to 0+, lim f( (x)) = 1, x tends to 0-, lim f( (x)) = 1, f( (x)) is continuous at x=0.

2) f(x)=x, (x)=1 x (0)=1), x=0 is the infinite break point of (x), and f( (0)) =1

x tends to 0+, lim f( (x)) = + x tends to 0 -, lim f( (x)) = - x=0 is the f( (x)) infinite discontinuity.

3) f(x)=x, (x)=x 2 ( 0)=1), x=0 is the discontinuable discontinuity point of (x), and f( (0)) =1

x tends to 0+, lim f( (x)) = 0, x tends to 0-, lim f( (x)) = 0, x=0 is the discontinuity point of f( (x)).

4) f(x)=x, (x)= x+1 (x>0) ,x)= x-1 (x<0) ,0)= 0, x=0 is the hop break point of (x), f( (0)) =0, x tends to 0+, lim f( (x)) = 1, x tends to 0-, lim f( (x)) = -1, x=0 is the hop break point of f( (x)).

From the above, it can be seen that the definition domains of (x), f(x) are all (- f(x) is continuous on (- , x) has a discontinuity x=x0, if f( (x) is discontinuous at x=x0, f( (x) and (x) have the same discontinuity type at x=x0.

2.x) is defined by (-a)(a,+ f( (x)) may be continuous.

1）f（x)= x ，φx ) =| x |x≠0) ,x) is defined by (-0) (0,+ range (0,+ f( (x)) is continuous on (0,+;

2) f(x)=x, (x)=1 x, x) is (-0)(0,+ f( (x)) is discontinuous at x=0.

3) f(x)=x, (x)=x(x≠1) ,x) is (-1)(1,+ f( (x)) is discontinuous at x=1.

1.f( (x)) may be continuous.

For example, f(x)=x, (x)=x,x≠0, where f( (x)) is discontinuous at x=0.

2.The answer is that f( (x)) may be continuous.

For example, f(x)=x, (x)=x,x>0, (x)=-x,x<0, since the (x) range is (0,+ the function image is the same as f( (x))=x,x>0.

x in the defined domain (-0) (0, + consecutive.

The answer is as follows: <>

For reference, please smile.

D is not necessarily correct ABC, D can use the counter-argument method If D is incorrect, i.e. D makes the continuous by; The four operations of the continuous function are based on the fact that f(x) is continuous and (x) is also continuous and contradicts the problem, so the assumption d is wrong, so there must be a break point in d.

Solution, the second, the fourth is right.

The first and third are wrong, if tanx is combined with arctanx, there is no discontinuity in x.

Find all the discontinuities first, then the interval between the discontinuities is a continuous interval.

x=0 is the hop break.

x=1 is the discontinuity point that can be removed.

x=2 is the infinite discontinuity.

0,1) and (1,2) are continuous intervals, but only (0,1) is bounded.

A type of discontinuity point is an isolated point where the function is undefined, but immediately on both sides of the point, and the function value (limit) is the same;

Other discontinuities are undefined solitary points of the function, which are immediately flanked by the point, and the values (limits) of the function are different.

1) Fraction, the point with a denominator of 0 is the discontinuity point.

y=(x-1)(x+1) (x-1)(x-2), x=1, x=2 are discontinuities, but if x≠1, x-1 can be approximated, y=(x+1) (x-2), as long as the definition is added, x=1, y=(x+1) (x-2), the function is continuous at x=1, and x=2 cannot be removed.

2) When x=k, tanx=0, the denominator is 0, is the break point, on both sides of the point, the value of tanx is close to 0, after the reciprocal, respectively, infinity, discontinuous, and can not go.

3) x tends to 0, 1 x approaches infinity, and the value of cosx is uncertain and, therefore, cannot go.

4) x approaches 1 from the left side, y approaches 0, x approaches 1 from the right side, and y approaches 2, different, can't go.

Seeing whether the left and right limits are the same is the basic way to judge whether it is possible to go.

Let a=f(x) be defined as follows.

f(x)= tan( x 2) if|x|>1, x does not belong to af(x) = tan( (2x)) if|x|<1, x does not belong to af(x) = 0 if x a can satisfy the requirement.

There are many such functions, for example, I will give you a random one, f(x)=1 sin(n*pi x), at the points you mentioned x=0, +(1,+(2,+(1 2,..n,+(1/n,..It's all discontinuity, and it's an infinite number of discontinuities.

pi=,n is any natural number. Hope it helps.

A: Define f(x) = tan[(2x+1) 2] (when x is an integer), tan[(1+2 x) 2] (when x is not an integer).

Then x is f(x) the second type of discontinuity when it meets the above conditions, and both are infinite discontinuities.

If you don't want to segment the function, this function can:

f(x)=1/sin(πx) +1/sin(π/x)

Fine, just get a piecewise function.

f(x)=g(x) x is not equal to 0,+(1,+(2,+(1 2,..n,+(1/n,..

f(x)=0 x=0,+(1,+(2,+(1/2,..n,+(1/n,..

x is close to 0+, x is the positive infinitesimal, a 0, x a infinitesimal , x a is the infinitesimal reciprocal, very large, sinx is **.

In fact, the purpose of the classification discussion is to make it clear that different values produce different results: the limit of 0+ at 0 is 0

0 is an oscillation.

So the results will be different.

This function is continuous from 0 to 2.

First of all, f(x) is not defined at x=0, and in order for the function to be continuous at that point, the limit of the function at that point must be equal to the defined value of the function.

When x tends to 0, cotx 1 tanx 1 x (equivalent infinitesimal relation) then f(x)=(1-x) (1 x), and -x as t, then f(t)=(1+t) (1 t).

Because the important limit (1+t) (1 t) = e when t tends to 0, so f = 1 e when t tends to 0

Then when x tends to 0, f(x) tends to 1 e