Find out how to determine the number of discontinuities... 5

You can go to the break point: The function exists and is equal to the left and right limits at the point, but is not equal to the function value at the point or the function is not defined at the point.

Hop break point: The function exists at the left and right limits of the point, but is not equal.

Reachable and hop breaks are called first-class breaks.

It is also called a finite discontinuity. Other discontinuities are called type 2 discontinuities.

The method is to find the left and right limits separately, and then judge whether it is possible to go or jump according to the definition of the point and the above two points, if not, it is the second type of discontinuity.

Type 2 discontinuity: At least one of the left and right limits of a function is absent.

a.If there is at least one infinity of the left and right limits of the function at x=xo, then x=xo is said to be an infinite break point of f(x). Example: y=tanx, x=2.

b If at least one oscillation of the left and right limits of the function at x=xo does not exist, then x=xo is said to be the oscillation break point of f(x). Example: y=sin(1 x), x=0.

Higher Mathematics is the study of elementary functions

Generally, the discontinuity point is found by observation

Grasp the undefined points of the function (e.g. points with a denominator equal to 0).

The segmentation point of the piecewise function

as well as commonly used discontinuities such as LNX, Tanx, etc

It's nothing more than a combination of them.

【Appendix】Definition of Intermediate Breakpoint in Advanced Mathematics:

If a function is not continuous at a certain point, that point is called a break point of the function.

According to this definition, the break point of a function is nothing more than three possibilities:

The function is not defined at that point;

The function has no limit at that point;

The function has a definition at that point and also a limit, but the limit ≠ the value of the function.

1, that is, x=5, the point that makes the function meaningless is the discontinuity point, where the denominator is not 0, that is, x-5 is not equal to 0, so x is not equal to 5. Other than this, because the numerator and denominator are all elementary functions and are continuous, there are no other breakpoints.

When x=1, the left limit (from negative infinity to 1) of the function is equal to , and the right limit (from positive infinity to 1) is equal to ; The left limit is not equal to the right limit and is the jump break point in the first type of break point.

When x = 1, the left limit of the function is equal to 0, and the right limit is equal to 0, but the function is meaningless at that point, so it is a removable break point in the first type of discontinuity.

Let the unary real function f(x) be defined in some decentered neighborhood of the point x0. If the function f(x) has one of the following things:

1) The function f(x) exists but is not equal at the left and right limits of point x0, i.e., f(x0+) ≠ f(x0-).

2) At least one of the functions f(x) does not exist in the left and right limits of the point x0.

3) The function f(x) exists and is equal to the left and right limits at the point x0, but is not equal to f(x0) or f(x) is undefined at the point x0.

Then the function f(x) is discontinuous at the point x0, and the point x0 is called the break point of the function f(x).

There is only one discontinuity, and the denominator cannot be zero!

So x≠5

The break point is the point where the denominator is 0, so that the denominator x(x 2-1)=0, then the break point x=-1,x=0,x=1, at x=-1, the left limit = the right limit = 3 2, is the discontinuity point;

At x, the limit = is an infinite break point.

f(x) =x-1) Zhishu [|x|(x 3-1)(x-2)] breakpoint = > shouting denominator infiltrator = 0

x|(x^3-1)(x-2) =0

x=0 or 1 or 2

3 breakpoints.

The formula for finding the discontinuity point is: y=ad*q. A discontinuity point is an interruption at xo at a point in the discontinuous function y=f(x), then xo is called the discontinuous point of the function.

Breaks can be divided into infinite and non-infinite breaks, and in non-infinite breaks, there are also go breaks and jump breaks.

A continuous function is a function y=f(x) that causes a small change in the independent variable x, which causes a small perturbation of the dependent variable y. For example, the temperature changes over time, and as long as the time changes very little, the change in temperature is also very small; Another example is the change of displacement from the falling body with time, as long as the time change is short enough, the change in displacement is also very small. For this phenomenon, the dependent variable is continuously variable with respect to the independent variable, and the image of the continuous function in the Cartesian coordinate system is a continuous curve with no breaks.

The discontinuity point can be removed, that is, the left limit = right limit = finite value, which has nothing to do with the value of this point and the presence or absence of definition, and can be redefined to make it continuous.

The "finite point" (not counting x) with a 0 parent may be discontinuous, so take it out and discuss it in turn. x=0, x=-1, and x=1

1) When x base group 0, because it involves |x|, so it is necessary to discuss it on both sides.

When x 0+, limf(x)=lim(x x-1) [x(x+1)lnx]=lim(x x-1) (xlnx).

Since 0 0 = 1, the numerator is close to 0 at x 0+; For the denominator, xlnx=lnx (1 x), l is applied'The hospital rule is known to be close to 0 at x 0+.

Therefore, the numerator denominator satisfies the l of the 0 0 type'hospital rule, lim(x x-1) (xlnx) = lim(x x) (lnx+1) (lnx+1) = lim(x x) = 1

When x 0+, limf(x)=lim(x x-1) [x(x+1)lnx]=lim(x x-1) (xlnx).

In the same way, the numerator denominator satisfies the L of type 0 0'hospital rule, lim(x x-1) (xlnx) = lim(x x) (lnx+1) (lnx+1) = lim(x x) = 1

In summary, when x 0, the left limit = right limit = 1, therefore, x = 0 is the discontinuity point that can be removed.

2) When x -1, limf(x)=lim[(-x) x-1] [x(x+1)ln(-x)]=lim[1-(-x) x] [x+1)ln(-x)].

The situation is similar to x 0, molecule 1-(-x) x 0;The denominator (x+1) ln(-x) satisfies l'The hospital rule, the limit of which is 0

So, in general, the L of the 0 0 type is satisfied'hospital rule, limf(x)=lim[1-(-x) x] [x+1)ln(-x)]=lim[-(x) x][ln(-x)+1] [ln(-x)+(x+1) x].

Among them, x -1+ is + and x -1-, which is an infinite break point and does not meet the requirements. Abandon it.

3) When x 1, limf(x)=lim(x x-1) [x(x+1)lnx]=lim(x x-1) (2lnx).

The numerator denominator satisfies the L of the 0 0 type'Hospital rule, there is.

lim(x x-1) (2lnx)=lim(x x)(lnx+1) (2 x)=1 2, so x=1 is also a discontinuity point.

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The formula for finding the discontinuity point: e (1 ) e=01 x. The discontinuity point refers to the interruption phenomenon at a certain point at xo in the non-continuum fiber continuation function y=f(xa), then xo is called the discontinuous point of the function.

A continuous function is the function y=f(x) when the change in the independent variable x is small, the change in the dependent variable y is also small. For example, as long as the time change is small, the change in the imitation temperature of the air mass is also very small; Another example is that the displacement of a free-fall body varies with time, and as long as the time change is short enough, the change in displacement is also very small.

Two intermittent chaotic wide points: posture accompanies this.

x= 2x=1 (can be removed).

Denominator = x -x=x (x -1) = x (x + 1) (x-1).

where (x-1) is about separated from the molecule's (x-1), so:

f（x）=tanx/[x（x+1）]。

The domain of tanx is defined as: x≠k + 2;For f(x), x≠0, x≠-1.

x=1 is the beam dust circle at the discontinuity point;

x=0, x=-1, x=k + 2 are the discontinuities of the rubber collapse.

x=0,1, -1, denominator=0

f(x) =tanx. (x-1)/(x^3-x)lim(x->0) f(x)

lim(x->0) tanx.(x-1) number of ridges (x 3-x) lim(x->0) x (x-1)/[x.

x 2-1)] lim(x->0) (x-1) (x 2-1)x=0: Breakable (type 1 break).

lim(x->1) f(x)

lim(x->1) tanx.(x-1) (x 3-x) lim (x-> cha talk 1) tanx (x-1)/[x.(x-1)(x+1)]

lim(x->1) tanx [x(x+1)]x=1: can go to the break point (type 1 break point Sakura Keito) lim(x->-1+) f(x).

lim(x->-1+) tanx. (x-1)/(x^3-x)lim(x->-1) tanx. (x-1)/[x.(x-1)(x+1)]

Infinitude. x=-1: Infinite discontinuity (type 2 discontinuity).