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If f (x) has a limit at x = x0, then f (x) has a limit at x = x0 A may not be defined B continuous C-differentiable D discontinuity
A
Boundedness of functions
How to know if a function is bounded? How to judge?
If there is a positive number m in the range (represented by D) of the variable x, the value f (x) of the function on D satisfies
│f(x)│≤M ,
Then the function y = f (x) is said to be bounded on D, or F (x) is said to be bounded on D. if there is no such positive number m, then the function y = f (x) is said to be unbounded on D, or F (x) is said to be unbounded on D
give an example:
Generally speaking, a continuous function is bounded in a closed interval. For example, y = x + 6 has a minimum value of 7 and a maximum value of 8 on [1,2], so its function value changes between 7 and 8, which is bounded, so it is bounded
What does bounded function mean
The function is bounded, which means that the graph is framed between two straight lines parallel to the X axis from the geometric point of view, and will not run out; from the algebraic point of view, it means that the value of the function will not tend to positive infinity or negative infinity; at that time, it does not mean that there is a limit, such as y = SiNx, which is framed between the two straight lines y = ± 1. When x →∞, SiNx swam between [- 1, + 1]
Prove whether the function is bounded or not
y=(x6+x4+x2)/(x6+1)
When x is infinite, the denominator of LIM y is divided by x ^ 6
=lim (1+1/x^2+1/x^4)/(1+1/x^6)
=1.
From the boundedness of limit, we know that there exists M. when | x | > m, there must be
|y-1|
Let vector a plus vector b be equal to (4, - 2) vector a minus vector 2B be equal to (1, - 8) find the sine value of the angle between vector 2a and a-b
A + B = (4, - 2) a-2b = (1, - 8) → B = (1,2), a = (3, - 4), → 2A = (6, - 8), A-B = (2, - 6) the cosine value of their angle is 2A (a-b) / l2all (a-b) l = 3 times root 10 of 10, and its sine value is 10 times root 10 of 10
Why is sine 90 degrees equal to 1
Draw a picture to illustrate
Because this is a theorem, we should use it flexibly
What is the sine of 120 degrees, 180 degrees?
Sin120 = sin60 = two thirds root sign three
sin180=sin0=0
There's a formula that assumes that angle a is between 0 and 180
Then Sina = sin (180-a)
What is 1-cosx equal to? Sine
Thank you for your reply. Can you tell me which formula you derived it from
cosx=(cosx/2)^2-(sinx/2)^2=1-2(sinx/2)^2
1-cosx=2(sinx/2)^2
In a triangle, a, B and C are the opposite sides of angles a, B and C respectively. Let a + C = 2B and a-c = Pai / 3, find the sine value of angle B
Because a + C = 2B
From the sine theorem, we can know that sina + sinc = 2sinb
From the sum difference formula of integration
sinA+sinC=2* sin[(A+C)/2]* cos[(A-C)/2]
Because a + B + C = 180 degree, a-c = 60 degree
therefore
sinA+sinC=2* sin[(A+C)/2]* cos[(A-C)/2]
=2*sin(90°-B/2)*cos30°
=√3cos(B/2) ②
It can be concluded from two formulas
2sinB=√3cos(B/2)
SINB = 2Sin (B / 2) * cos (B / 2)
therefore
4sin(B/2)*cos(B/2)=√3cos(B/2)
So sin (B / 2) = √ 3 / 4
Because B / 2 must be an acute angle,
So cos (B / 2) = √ 13 / 4
therefore
sinB=2sin(B/2)*cos(B/2)=√39/8
If the cosine of three angles of a triangle is the sine of three angles of another triangle, what triangle is it
An acute triangle
The three sine values of the other triangle are greater than 0 and less than or equal to 1 [the sine value of 0 degree or 180 ° angle is 0, the sine value of the triangle triangle is not zero, and zero does not constitute a triangle]
The cosine value of a triangle is greater than - 1 and less than - 1 [0 degree cosine is 1180 ° cosine is - 1, the cosine value of a triangle is not - 1, or 1, is - 1, or 1 does not constitute a triangle]
Therefore, the cosine values of the three angles of the triangle are greater than zero and less than 1, that is, there is no right angle or obtuse angle
So it's an acute triangle
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