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What is the limit of sinx-x / x ^ 3 approaching zero?
The limit is of 0 / 0 type, and the first use of Robida's law
lim(sinx-x)/x^3
=lim（cosx-1)/3x^2
=lim(-x^2/2)/3x^2
=-1/6
The infinitesimal equivalent substitution cosx-1 = - x ^ 2 / 2 is used
Given that the circle x ^ 2 + y ^ 2 + 8x-6y + 21 = 0 and the line y = MX intersect at two points P and Q, O is the origin of the coordinate, find the product of the vector OP and the vector OQ
Y = MX is substituted into equation x square + y square + 8x-6y + 21 = 0 x ^ 2 + m ^ 2x ^ 2 + 8x-6mx + 21 = 0 (1 + m ^ 2) x ^ 2 + (8-6m) x + 21 = 0 x1x2 = 21 / (1 + m ^ 2) P (x1, MX1) Q (X2, MX2) | op | = √ (1 + m ^ 2) | x1 | OQ | = √ (1 + m ^ 2) | x2 | op | OQ | = (1 + m ^ 2) | x1 | OQ | = 21
The limit when x ^ SiNx, X tends to zero
Let y = x ^ SiNx. Take logarithm on both sides to get LNY = SiNx * LNX. (1). It is easy to know that when x --- > 0, SiNx * LNX is of 0 * ∞ type. According to the law of lobita, SiNx * LNX = (LNX) / [1 / SiNx] = (1 / x) / [- cosx / (SiNx) ^ 2] = - [(SiNx) / x] * TaNx --- > - 1 * 0 = 0. (X -- > 0). (Note: the above equality refers to limit equality)
We can first calculate the value a of LNX ^ SiNx, and then the limit is e ^ a
The limit of SiNx LNX = LNX / (1 / SiNx) is 0
So the answer to the question is 1
Limx ^ SiNx = lime ^ (sinxlnx) = lime ^ (xlnx) = e ^ (limxlnx) let t = - LNX, then limxlnx = - limt / e ^ t (t tends to positive infinity) = 0, so limx ^ SiNx = e ^ 0 = 1
lim x^sinx
=lim e^(sinxlnx)
=e^lim sinxlnx
=e^lim lnx/(1/sinx)
=E ^ LIM (1 / x) / (- cosx / sin ^ 2 x) (Law of lobita)
=e^lim (-sin^2 x/xcosx)
=e^0
=1
If n (x, y) satisfies the inequality system, X is greater than or equal to 1
Given that O is the origin of coordinates and m (- 2,0), if n (x, y) satisfies the inequality system, X is greater than or equal to 1, y is greater than or equal to 0, and X + y is less than or equal to 4, then the value range of vector OM * on is, and the answer is [- 8,1}
Sorry, point m is (- 2, 1)
OM*ON=-2x
Because x is greater than or equal to 1, y is greater than or equal to 0, and X + y is less than or equal to 4
So x is greater than or equal to 1 and less than or equal to 4
So om * on = - 2x is greater than or equal to - 8 and less than or equal to - 2
Are you writing the title wrong
The vector om * on is a complex number
It is necessary to write the value range of real part and imaginary part respectively
I'm sorry, point m is (- 2,1)
Does SiNx have a limit when x tends to zero
There exists, which is equal to 0, because sin is a continuous function, so Lim SiNx = sin0 = 0
Let point o be the origin of coordinates, m (2,1) if point n (x, y) satisfies the inequality system: ① x-4y + 3 ≤ 0; ② 2x + Y-12 ≤ 0; ③ x ≥ 1,
Then the maximum value of the modulus * cos angle mon of the vector on is
This is a linear programming problem. From the drawing, we can see that the feasible region is a triangle composed of inequality system. Om | on = | om | on | cos ∠ Mon, so | on | on | mon = (2x + y) / √ 5. Because the objective function | on | mon = (2x + y) / √ 5 and
How to find the limit of (2 ^ x-1) / X when x approaches 0?
Don't just be the result, there are a few steps
[Law of Robita]
lim(x->0) (2^x-1)/x
=lim(x->0) ln2 * 2^x /1
= ln2
[equivalent infinitesimal]
Let: 2 ^ X - 1 = t, then: x = ln (1 + T) / LN2, X - > 0, T - > 0, ln (1 + T) ~ t
lim(x->0) (2^x-1)/x
=lim(x->0) t/[ln(1+t)/ln2]
=lim(x->0) ln2 t/ln(1+t)
= 1
[important limit]
Let: 2 ^ X - 1 = t, then: x = ln (1 + T) / LN2, X - > 0, T - > 0
lim(x->0) (2^x-1)/x
=lim(x->0) t/[ln(1+t)/ln2]
=lim(x->0) ln2/ln[(1+t)^(1/t)]
= ln2/lne
= ln2
Let m (a, b) be in the plane region determined by the system of inequalities x ≥ 0, y ≥ 0, x + y ≤ 2, then the maximum distance from n (a + B, a-b) to the origin of the coordinate
The original system of inequalities only plays the role of new conditions after transformation,
Let x = a + B, y = A-B, then a = (x + y) / 2, B = (X-Y) / 2
Because a and B are in the region, there is a new system of inequalities a ≥ 0, B ≥ 0, a + B ≤ 2
Substituting a and B into a new inequality of X and Y
Draw an isosceles right angle triangle whose base height is 2, and know the root of the maximum distance is 2
Root 8 2 times root 2
My idea is that a + B is less than or equal to 2, and the maximum is 2. When A-B is the maximum, a is 2, B is 0, and 2 n (2, 2) to the origin is root 8
When x approaches 0, what is the limit of [(1 + x) / (1-x)] ^ x?
Given that the point P (x, y) satisfies the linear constraint condition y ≤ 2x + y ≥ 1x − y ≤ 1, the point m (3, 1) and o are the coordinate origin, then the maximum value of OM · OP is ()
A. 12B. 11C. 3D. -1
Let z = om · OP, then z = 3x + y, i.e. y = - 3x + Z, make the plane region corresponding to the inequality system, as shown in the figure: translate the straight line y = - 3x + Z, from the image, when the straight line y = - 3x + Z passes through point a, the intercept of the straight line y = - 3x + Z is the largest, at this time Z is the largest, from y = 2x − y = 1, the solution is x = 3Y = 2, i.e. a (3,2), at this time z = 3x + y = 3 × 3 + 2 = 11, so the maximum value of OM · OP is 11, so select B
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