Prove the boundedness of function f (x) = x / (x ^ 2 + 1)

Prove the boundedness of function f (x) = x / (x ^ 2 + 1)

A simple problem of proving function bounded
The function f (x) is continuous in (a, b), and the limit of F (x) at a and the left limit at B exist. It is proved that the function is bounded on (a, b)

Because f (x) has a right limit at a, we know that there exists δ 1 > 0 according to the properties of the limit, so that the interval (a, a + δ 1) is bounded
Because f (x) has a left limit at B, we know that there exists δ 2 > 0 according to the properties of the limit, so that the interval (B - δ 2, b) is bounded
For any 0

The proof of function boundedness
Prove that the function y = (x + 1) / (x Λ 4 + 1) is bounded, find the proof method!

When | x|

What is the boundedness of functions

The boundedness of a function refers to the finiteness of the value range of a function. For example, the sine function f (x) = SiN x, whose value range is - 1 to 1, is a limited range, so we can say that the function is bounded, while the value range of y = x is R, which is an infinite range, so we can say that the function is unbounded

If the three sides of a triangle are known to form an arithmetic sequence with a tolerance of 2 and the sine value of its maximum angle is 32, then the area of the triangle is______ ．

Let three sides be A-2, a, a + 2 (a > 0), then a + 2 is the largest side. According to the angle of the triangle, the angle of the pair is the largest angle ∵ the sine value of the largest angle is 32, then the largest angle is 120 ° according to the cosine theorem, cos120 ° = (a − 2) 2 + A2 − (a + 2) 22a (a − 2) = - 12, a2-5a =

The angle a sine is greater than the angle a sine =?
Please write down the detailed steps

Sin a = 2Sin (A / 2) * cos (A / 2) so sin (A / 2) / sin a = 1 / {2cos (A / 2)}

The sine of a base angle of an isosceles triangle is 5 / 13, then the sine of the top angle of the triangle
If the sine value of a base angle of an isosceles triangle is 5 / 13, then the sine value of the apex angle of the triangle is

Let the isosceles triangle AB = AC
sinB=sinC=5/13
cosB=cosC=12/13
sinA=sin(B+C)
=sin2B
=2sinBcosB
=2*(5/13)*(12/13)
=120/169
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As shown in the figure, in square ABCD, O is a point on the edge of CD, and the semicircle with o as the center and OD as the radius is just circumscribed with the arc of the sector with B as the center and BC as the radius, then the sine value of ∠ OBC is___ ．

Let the side length of the square be 1 and the radius of the semicircle be X. then ob = 1 + X, OC = 1-x. in RT △ OBC, according to the Pythagorean theorem, we obtain (1 + x) 2 = (1-x) 2 + 1, x = 14. Then ob = 54, OC = 34. Then sin ∠ OBC = ocob = 35

What is the sine cosine of 0, π / 6, π / 4, π / 3, π / 2, 2 / 3 π, 3 / 4 π, 5 / 6 π, π, 7 / 6 π?

sin0=0sinπ/6=1/2sinπ/4=√2/2sinπ/3=√3/2sinπ/2=1sin2/3π=sin(π-1/3π)=sin1/3π=√3/2sin3/4π=sin(π-1/4π)=sin1/4π=√2/2sin5/6π=sin(π-1/6π)=sin1/6π=1/2sinπ=0sin7/6π=sin(π+1/6π)=-sin1/6π...

By using the sine cosine function values of units 0. π / 6, π / 4, π / 3, π / 2, the sine cosines of which angles in the interval (- 7 π / 5,3 π / 5) can be obtained
Function values? And these are called sine cosine function values

By using the sine cosine function values of unit 0. π / 6, π / 4, π / 3, π / 2, we can find out which angle's sine cosine function values in the interval (- 7 π / 5,3 π / 5)] analytic: ∵ interval (- 7 π / 5,3 π / 5)] sin (- 4 π / 3) = - sin (4 π / 3) = sin (π / 3) = √ 3 / 2, cos (- 4 π / 3) = - 1 / 2Sin (- 5 π / 4) = √ 2 / 2