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When x approaches zero, is the limit of Ln (x / SiNx) x / sinx-1 Or how to find the limit of 1 / xsinx power when x approaches 0 (xcosx / SiNx)
When x approaches 0, the limit of Ln (x / SiNx) is 0,
When x approaches 0, the limit of X / SiNx is 1,
So the limit of Ln (x / SiNx) is 0
It is necessary to take the process of - 178
=-(3x²-4y³)(3x²+4y³)-(3x²+4y³)²
=-(3x²+4y³)(3x²-4y³+3x²+4y³)
=-6x²(3x²+4y³)
=-The fourth power of 18x - 24x & # 178; Y & # 179;
F (x) = (LN | x | / | X-1 |) SiNx why is the limit zero when x approaches zero?
First of all, the term X-1 is not important, because when X - > 0, it has a limit of 1. Sin (x) and X are infinitesimals of the same order, as long as X * ln | x | tends to 0. We can directly use the law of lobita: Lim x * LN | x | = LIM (LN | x |) '/ (1 / x)' = lim - (1 / x) / (1 / x ^ 2) = lim - x = 0, or we can see this way: let x = 1 / T
Is ln | x | and x ^ 2 infinitesimals of the same order? Who said that?
The distance between the center of circle 5x & # 178; + 5Y & # 178; + 20x-10y + 24 = 0 and the line 3x + 4Y = 0 is
5x²＋5y²＋20x－10y＋24=0
x²＋y²＋4x－2y＋24/5=0
(x+2)^2＋(y-1)^2-4-1-24/5=0
The center of the circle is (- 2,1) according to the distance formula from the point to the straight line
│【3*（-2）+4*1+0】│/(3^2+4^2)^0.5=2/5=0.4
5x²＋5y²＋20x－10y＋24=0
(x + 2) ² + (Y-1) ² = 1 / 5 = > Center (- 2,1)
The line passing (- 2,1) and perpendicular to 3x + 4Y = 0 = > y = (4 / 3) x + 11 / 3
Intersection (- 44 / 25, 33 / 25)
Distance (root 14) / 25
How to find the limit of (tanx-x) \ \ (x-sinx) when x approaches 0, please explain in detail the limit of (arctanx-x) \ \ ln (1 + X & sup3;) when x approaches 0
The law of lobida is enough
5X/3+4Y=10.4 ,3X/4+0.5Y=39/20
Addition subtraction elimination method
5X/3+4Y=10.4 (1)
3X/4+0.5Y=39/20 (2)
(1)-(2)*8
5X/3+4Y-6X-4Y=10.4-15.6
-13X/3=-5.2
X=1.2
Y=(10.4-5X/3)/4=2.1
The limit of x-arcsinx / ln (1 + SiNx & # 179;) when x tends to zero
The distance from the center of the circle X & sup2; + Y & sup2; - 10Y = 0 to the straight line 3x + 4y-5 = 0 is the detailed solution process
Circle x ^ 2 + (Y-5) ^ 2 = 25
Center (0,5)
d=(3*0+4*5-5)/√（3^2+4^2)=3
The standard equation is x ^ 2 + (Y-5) ^ 2 = 25
The center of the circle is (0,5)
From the formula, d = | 3 × 0 + 4 × 5-5 | / radical (3 ^ 2 + 4 ^ 2) = 3
Note: the formula of the distance from the center of the circle (x, y) to the straight line ax + by-c = 0 is d = | ax + by-c | / radical (a ^ 2 + B ^ 2)
The limit when (TaNx SiNx) △ Sin & # 178; X △ ln (1 + x) x tends to zero
Ax square + ay square-4 (A-1) x + 4Y = 0 represents the general equation of the circle. Find the value range of the real number a and the equation of the circle with the smallest radius
x²+y²-4(a-1)x/a+4y/a
formula
[x-2(a-1)/a]²+(y+2/a)²=[2(a-1)/a]²+(2/a)²
Then (Supa; - 2) = (Supa; - 2) / Supa
(a-1)²/a²+1/a²>0
Then as long as a & sup2; ≠ 0
So a ≠ 0
r²=4[(a-1)²/a²+1/a²]
=4(a²-2a+2)/a²
Let y = (A & sup2; - 2A + 2) / A & sup2;
(1-y)a²-2a+2=0
The discriminant is greater than or equal to 0
4-8(1-y)>0
y>=1/2
So (A & sup2; - 2A + 2) / A & sup2; minimum = 1 / 2
2a²-4a+4=a²
A=2
So x & sup2; + Y & sup2; - 2x + 2Y = 0
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