In the problem of higher numbers, find the limit of 2x cube + X part [(x square + 1) times SiNx] when x approaches 0

In the problem of higher numbers, find the limit of 2x cube + X part [(x square + 1) times SiNx] when x approaches 0

 
The first limit is obviously 0; the second is the product of infinitesimal and bounded quantity, so it is also 0; the third is the important limit
Document: polynomial by polynomial calculation (3x-2y) (5x + 3Y)
(3x-2y) (5x + 3Y) = 3x (5x + 3Y) - 2Y (5x + 3Y) = 3x × 5x + 3x × 3y-2y × 5x-2y × 3Y = 15x ^ 2 + 9xy-10xy-6y ^ 2 = 15x ^ 2-xy-6y ^ 2. For multiplication of two binomials, the formula is: front, back, inside, outside, that is: front multiplied by front + back multiplied by back + inside multiplied by inside
(a-b)(c+d)
=ac+ad-bc-bc
So = 15x + 9xy-10xy-6y
=15x-xy-6y
15X"-XY-6Y"
Find the limit of (SiNx / x) ^ (1 / x ^ 2) when x approaches 0
See if you're satisfied & nbsp; thank you
Given that the polynomial - 3x ^ ay ^ 2-1 / 5x ^ 3Y ^ 2 + 1 / 3x ^ 2Y is a sixth degree polynomial, find the value of (7-2a) ^ 2012
It's a polynomial of degree six
So the first number is a + 2 = 6
A=4
So the original formula = (- 1) ^ 2012 = 1
How to prove that the limit of SiNx when x approaches 0 is 0?
Zero
When x approaches 0 +, limsinx = 0
When limsinx = 0, it tends to zero
The left and right limits are equal
So Lim SiNx = 0 (when x tends to 0)
Drawing proof
dsinx=cosx,x-0,cosx=1,x-0
Just draw a curve of sin
Since SiNx is a continuous function and sin0 = 0, the original proposition holds.
It is equivalent to LIM (x → 1) x + 1 = 2
The original question is wrong! How to prove SiNx when x approaches 0, the limit is 1? This is a problem of finding the derivative, because the derivative SiNx is cosx, so when x approaches 0, substituting 0 into cosx will get 1
Let f (x) = SiNx proof: because: | f (x) - a | = | sinx-sin0 | = | SiNx | & lt; | x | - SiNx | & lt; | x |, please draw the unit circle: the vertical distance from the point to the straight line is the smallest, so SiNx is the shortest distance from the point on the circle to the X axis. |X | is the arc length. &In order to make | f (x) - a | & lt; | x | & lt; E, for any given positive number e, we can take e = Q. when x is suitable for the inequality: 0 & lt; | x | & lt; Q, the corresponding output value f (x) satisfies the inequality: | f (x) - sin0 | = | SiNx | & lt; Q, thus & nbsp; limsinx & nbsp; X tends to 0 & nbsp; = 0
Given that a is in the plane region represented by the inequality system X − 2 ≤ 03x + 4Y ≥ 4Y − 3 ≤ 0, and point n is on the curve x2 + Y2 + 4x + 3 = 0, then the minimum value of | an | is______ .
First, the feasible region is drawn according to the constraint conditions, z = | an |, ∵ the distance between the point in the feasible region and the center of the circle B (- 2, 0). When the distance between the point B and the straight line 3x + 4y-4 = 0, Z is the minimum, the minimum value is 2, and the minimum value of | z = | an | = 2-1 = 1, so fill in: 1
When x approaches 0, find the limit of "1 / SiNx - 1 / X"
lim(1/sinx-1/x)=lim(1/sinx)-lim(1/x)
=LIM (1 / x) - LIM (1 / x) {because when x approaches 0, 1 / SiNx approaches 1 / x, the theorem of equivalent infinitesimal}
=0
Zero, right
lim1/sinx-1/x
=lim(x-sinx)/(xsinx)
=lim(x-sinx)/x^2
=lim(1-cosx)/(2x)
=lim(sinx)/2
=0
If the point P is in the plane region represented by the inequality system 2x-y + 2 > = 0, x + Y-2 = 0, then the range of the distance from P to the origin is
A. [radical 13 / 4, radical 10 / 2] B. [radical 13 / 4,2] C. [radical 10 / 2,2] D. [1 / 2,2]
Take D, and make the intersection area of three lines in the coordinate system. It is easy to find that the minimum value is 1 / 2, the maximum value may be 2, or the point where the line x + Y-2 = 0 intersects. By combining the above two line equations, we can see that the point is (3 / 2,1 / 2), and the distance from the point to the origin is: root 10 / 4, which is less than 2, so the value range is [1 / 2]. Over
You first find the intersection of three straight lines, a (- 3 / 4,1 / 2), B (3 / 2,1 / 2), C (0,2); then you can see from the graph that the nearest point to the origin is a, and the farthest point is C, and the answer is B
Find out the intersection of three straight lines, and then calculate the distance to the origin.
How to find the limit of (SiNx) / x? (x approaches 0)
Substitution by equivalent infinitesimal
At 0, SiNx ~ X
So, their limit is 1
In the chapter of limit of higher numbers, we have discussed the important limit in detail
Lim [sin (x) / x] = 1 (the limit process is x → 0)
The following conclusion is wrong: A. inequality X-Y + 1
C. The plane region below the line y = 1 can be expressed by the inequality y + 1
A. Inequality X-Y + 1
C