The limit of 1 / ln (x + 1) - 1 / SiNx when x tends to zero

The limit of 1 / ln (x + 1) - 1 / SiNx when x tends to zero

lim1/ln(x+1)-1/sinx
=lim [sinx-ln(x+1)]/sinx*ln(x+1)
=lim[sinx-ln(x+1)]/x*x
=lim(cosx-(1/x+1))/2x
=lim(-sinx+1/(x+1)^2)/2
=1/2
The equation AX2 + ay2-4 (A-1) x + 4Y = 0 denotes a circle. Find the value range of a and the equation of the circle with the smallest radius
(1) When ∵ a ≠ 0, the equation is [X-2 (a − 1) a] 2 + (y + 2a) 2 = 4 (A2 − 2A + 2) A2, because a2-2a + 2 = (A-1) 2 + 1 ＞ 0 is tenable, when ∵ a ≠ 0 and a ∈ R, the equation is a circle. (2) ∵ R2 = 4 · A2 − 2A + 2A2 = 4 (2A2 − 2A + 1) = 4 [2 (1a-12) 2 + 12], when ∵ a = 2, rmin2 = 2
How to find the limit of (E Λ x-e Λ SiNx) / (LN (SiNx Λ 3) + e Λ x) - x when x tends to 0?
The denominator should be ln (Sin & # 179; X + e ^ x) - X
The equation AX2 + ay2-4 (A-1) x + 4Y = 0 denotes a circle. Find the value range of a and the equation of the circle with the smallest radius
(1) When ∵ a ≠ 0, the equation is [X-2 (a − 1) a] 2 + (y + 2a) 2 = 4 (A2 − 2A + 2) A2, because a2-2a + 2 = (A-1) 2 + 1 ＞ 0, when ∵ a ≠ 0 and a ∈ R, the equation is a circle. (2) ∵ R2 = 4 · A2 − 2A + 2A2 = 4 (2A2 − 2A + 1) = 4 [2 (1a-12) 2 + 12], when ∵ a = 2, rmin2 = 2
How to find the limit of limx →∞ (x + cosx + 1 / x + SiNx + 2)?
Cosx and SiNx are bounded functions
So when x →∞
They can be omitted, so the limit is 1
limx→∞（x+cosx+1)/(x+sinx+2）
=limx→∞（1+cosx/x+1/x)/(1+sinx/x+2/x）
=(1+0+0)/(1+0+0)
=1
Note: limx →∞ (cosx / x) = limx →∞ (1 / X * cosx) = 0;
Limx →∞ (SiNx / x) = limx →∞ (1 / X * SiNx) = 0!!! Thank you very much.
If the circle (x + 3) square + (y-4) square = 16 is separated from the straight line x-ay-5 = 0, find the value range of real number a
Circle (x + 3) square + (y-4) square = 16
Center (- 3,4) radius r = 4
Away from the straight line x-ay-5 = 0
Then the distance from the center of the circle to the straight line d = i-3-4a-5i / √ (1 + A & # 178;) > R = 4
That is (- 8-4a) &# 178; > 16 (1 + A & # 178;)
16a²+64a+64-16a²-16>0
64a+48>0
The solution is a > - 3 / 4
If there are two different intersections between L1 and circle, the distance from the center of circle to the straight line is less than the radius. D = | 3k-4-k | / √ K ^ 2 + 1 & lt; 4. The solution is K & gt; 0 or K & lt; - 4 / 3. The discriminant method can also be used to change the straight line equation into oblique equation
Center of circle (- 3,4), radius = 4
From junior high school geometry knowledge
When a circle is away from a straight line, the distance from the center of the circle to the straight line should be greater than the radius
It can be seen from the distance formula from point to line
|-3-4a-5|/√(1+a²)＞4
That is | a + 2 | > √ (1 + A & # 178;)
The solution is a ＞ - 3 / 4
∴a∈(-3/4, +∞)
Find limit limx → infinity (SiNx ^ 2-x) / [(cosx) ^ 2-x]
Big head. The answer in the book seems to be - 1.
When x tends to infinity=
If the point (1, radical 3) is outside the circle x square + y square - 2aX - (2 radical 3) * ay = 0, find the value range of real number a
(x-a)²+(y-√3a)²=a²+3a²
The distance from the outside to the center (a, √ 3a) is greater than the radius
So (1-A) &# 178; + (√ 3 - √ 3a) &# 178; > A & # 178; + 3A & # 178;
4a²-8a+4>4a²
A0
So a
A is less than 1 / 2 and not equal to 0
Circle x square + y square - 2aX - (2 radical sign 3) * ay = 0
>
X ^ 2 - 2aX + A ^ 2 - A ^ 2 + y ^ 2 - (2 radical 3) * ay + 3A ^ 3 - 3A ^ 3 = 0
>
(x-a) ^ 2 + (y-radical (3) a) ^ 2 = (2a) ^ 2
The center of the circle is (a, radical (3)), and the radius is 2A
If the point is outside the circle, the distance from the point to the center of the circle is greater than the radius
So there are
X ^ 2 - 2aX + A ^ 2 - A ^ 2 + y ^ 2 - (2 radical 3) * ay + 3A ^ 3 - 3A ^ 3 = 0
>
(x-a) ^ 2 + (y-radical (3) a) ^ 2 = (2a) ^ 2
The center of the circle is (a, radical (3)), and the radius is 2A
If the point is outside the circle, the distance from the point to the center of the circle is greater than the radius
So there are
(1-A) ^ 2 + (radical (3) - radical (3) a) ^ 2 > 4A ^ 2
>
4*(1-a)^2 > 4a^2
=>1-a > a
perhaps
1-A
A < 1 / 2
Limx tends to clique, SiNx / clique-x, seeking limit
 
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The solutions of the equations 2x-m-2 = 0 and M-3 = 4Y of X satisfy the following conditions: 2x + y = 0, M =?
The solutions of the equations 2x-m-2 = 0 and M-3 = 4Y of X satisfy 2x + y = 0
2x=m+2
y=m/4-3/4
m+2+m/4-3/4=0
5m/4=-5/4
m=-5
Change the two formulas in the question
2x=m+2
y=(m-3)/4
2x+y=m+2+(m-3)/4=0
5m/4=-5/4
m=-1