Find (e ^ x-1) / X when x approaches the limit of 0,

Find (e ^ x-1) / X when x approaches the limit of 0,

Because when X - > 0
e^x-1->0 x->0
So we apply lobita's law, that is, to seek the derivatives of the denominator
limx->0(e^x-1)/x
=limx->0(e^x-1)'/x'
=limx->0 e^x/1
=1/1
=1
By using the law of lobida, the derivatives of numerator and denominator are 1
If you haven't studied lobida, expand e ^ x Taylor
By using the law of Robida, the original formula = e ^ X / 1, when x is close to 0, the value is 1
When x approaches 0, (e ^ x-1) and X are equivalent infinitesimals, so the limit of e ^ x-1) / X when x approaches 0 is 1
In the rectangular coordinate system, if O is the origin of the coordinate, point a (1,1) is known, and point P is determined on the x-axis so that △ AOP is an isosceles triangle, then there are several points P that meet the conditions
A. Four B, three C, two D, one. Notice that it's on the x-axis.
12. Think for yourself
B. Three, one point coincides
It's six. Think for yourself
The coordinates of P points are (1,0) (2,0) (- √ 2,0)
B. 3
There are three situations,
There are two solutions,
One has a solution
One is not true
（1,0）（2,0）（-√2，0）
When x tends to infinity, (x / (x + 1)) ^ x finds the limit
lim(x/(x+1))^x
=lim1/【(x+1）/x)】^x
=lim1/(1+1/x))^x
=1/e
Original formula = LIM (x - > ∞) [1-1 / (x + 1)] ^ x
=lim(x->∞) {[1+1/(-x-1)]^(-x-1)}^(-1)*[1+1/(-x-1)]^(-1)
=1/e
In the rectangular coordinate system, if O is the origin of the coordinate and a (1,1) is known, the point P is determined on the x-axis so that △ AOP is an isosceles triangle, then the number of qualified points P has ()
A. 6 B. 5 C. 4 d. 3
(1) If Ao is the waist, there are two cases: when a is the vertex, P is the intersection of a and the x-axis; when o is the vertex, P is the intersection of O and the x-axis; (2) when OA is the bottom, P is the intersection of the x-axis and the perpendicular of OA. The above four intersections do not coincide There are four points that meet the conditions
X tends to infinity. The limit of √ x (√ (x + 1) - √ x)
lim(x→0)√x[√(x+1) - √x]
　　= lim(x→0)√x/[√(x+1) + √x]
　　= lim(x→0)1/[√(1+1/x) + 1]
　　= 1/2
In the rectangular coordinate system, if O is the origin of the coordinate and (1,1) is known, a point P is determined on the x-axis so that △ AOP is an isosceles triangle, then there are several conditions for P to be satisfied
individual
Is (1,1) the coordinate value of a?
If you say (1, 1) If it is a point, because it is a triangle, AOP should be an isosceles triangle, which means: Ao = PO or AP = PO or AO = AP, and there is only one solution for P in these three cases, so there are three qualified P's to be proved by the counter proof method; there can't be two right angles in a triangle. It seems that my calculation is wrong and misleading. The counter proof is to let angle A and angle B in a triangle ABC be right angles The angle a plus the angle B is equal to 180 degrees, so AC and BC are parallel, so there is no intersection between them, that is, point C does not exist
If you say (1,1) is a point, because it is a triangle, AOP should be an isosceles triangle, which means: Ao = PO or AP = PO or AO = AP, and there is only one solution for P in these three cases, so there are three questions for P that meet the conditions. Question: use the counter proof method to prove that there cannot be two right angles in a triangle
Lime ^ X-1 / 2, X tends to 0, find the limit, why can it finally become X / 2x
=lime^x-1 /2 x
=lim(e^0-1 /2 x)
=lim(1-1 /2 x)
=lim(2x/2 x-1 /2 x)
=lim(x/2x)
=1/2
At the end of the rectangular coordinate system, O is the origin of the coordinate, a (1,1). Determine the point P on the x-axis. If △ AOP is an isosceles triangle, then it is consistent with point P common?
Let's answer. It seems that there are three. It seems that there are four,
4
(-√2,0) (1,0) (√2,0) (2,0)
√ denotes the root sign
Finding limit Lim x-0 2x / SiNx
Do you mean 2x / SiNx when LIM (x → 0)?
According to the law of lhospital, the derivative of the numerator and denominator of 2x / SiNx is obtained
When the original formula = LIM (x → 0), 2 / cosx = 2
Remember that X and SiNx are small quantities of the same order
X and SiNx are equivalent infinitesimals. If you have to use a process, use lobida, as follows:
lim 2x/sinx = 2*lim x/sinx
=2*lim 1/cosx
=2*1
=2
(3x²-4y³)(﹣3x²-4y³)-(﹣3x²-4y³)²
(3x²-4y³)(﹣3x²-4y³)-(﹣3x²-4y³)²
=(4y²)²-(3x³)²]-(3x²+4y³)²
=16y^4-9x^6-9x^4-24x²y³-16y^6
=16y^4-9x^4-24x²y³-25y^6