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大于小于号怎么区分
高数一道定积分证明,试证:0≤∫0~1(积分限)sin(x^n)dx≤ 1/n+1
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因为在[0,1]内0
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Proofs: Derivative
Trig Functions
| Trig Functions)
cos(x) = -sin(x)
tan(x) = sec^2(x)
csc(x) = -csc(x)
sec(x) = sec(x)
cot(x) = -csc^2(x)
Proofs of Derivative of Trig Functions
sin(x) : algebraic Method
Given: lim(d-&0) sin(d)/d = 1.
= lim(d-&0) ( sin(x+d) - sin(x) ) / d
= lim ( sin(x)cos(d) + cos(x)sin(d) - sin(x) ) / d
= lim ( sin(x)cos(d) - sin(x) )/d + lim cos(x)sin(d)/d
= sin(x) lim ( cos(d) - 1 )/d & + & cos(x) lim sin(d)/d
= sin(x) lim ( (cos(d)-1)(cos(d)+1) ) / ( d(cos(d)+1) ) &
+ & cos(x) lim sin(d)/d
= sin(x) lim ( cos^2(d)-1 ) / ( d(cos(d)+1 ) &
+ & cos(x) lim sin(d)/d
= sin(x) lim -sin^2(d) / ( d(cos(d) + 1) &
+ & cos(x) lim sin(d)/d
= sin(x) lim (-sin(d)) * lim sin(d)/d * lim 1/(cos(d)+1) &
+ & cos(x) lim sin(d)/d
= sin(x) * 0 * 1 * 1/2 + cos(x) * 1 = cos(x) &
& & Q.E.D.
cos(x) : from the derivative of sine
This can be derived just like
sin(x) was derived or more easily from the result of
Given: ; .
cos(x) = sin(x + PI/2)
sin(x + PI/2)
(x + PI/2) (Set
u = x + PI/2)
= cos(u) * 1 = cos(x + PI/2) = -sin(x) & &
tan(x) : from the derivatives of sine and cosine
tan(x) = sin(x) / cos(x)
sin(x)/cos(x)
= ( cos(x)
sin(x) - sin(x)
cos(x) ) / cos^2(x)
= ( cos(x)cos(x) + sin(x)sin(x) ) / cos^2(x)
= 1 + tan^2(x) = sec^2(x) &
& & Q.E.D.
Reciprocals
cot(x) : from
derivatives of their reciprocal functions
1/sin(x) = (
sin(x) ) / sin^2(x)
= -cos(x) / sin^2(x) = -csc(x)cot(x)
1/cos(x) = (
cos(x) ) / cos^2(x)
= sin(x) / cos^2(x) = sec(x)tan(x)
1/tan(x) = (
tan(x) ) / tan^2(x)
= -sec^2(x) / tan^2(x) = -csc^2(x)
& & & Q.E.D.
& 2000-. All rights reserved.&&&&&.&&&& Please read our .各种高等数学 1证明一个函数导数存在得先证明这个导数在这个点连续吗2 夹逼定理 书上说x大于0 然后x小于0 x就等于0 不对吧 比如书上证明sin(x) 是大于和小于03求用微积分证明点到直线最短距离公式 Ax+By+c=0 D=根号下(x-xo)+(y-yo)1和3题细节解释
左边=∫[0,x] (2t^2+2t|t|) dtt&=0时,∫ t|t| dt= ∫ t^2 dt= t^3/3 + C = |t|t^2 / 3 + C,t&0时,∫ t|t| dt = -∫ t^2 dt = -t^3/3 + C = |t| t^2 / 3 + C所以原式=(2t^3/3 + 2|t| t^2 / 3) |[0,x] = (2t^2/3 * (t + |t|)) |[0,x] = 2x^2/3 * (x+|x|)∫[a,b] f(t) f'(t) dt = ∫[a,b] f(t) df(t) = (分部积分) =(f(t) * f(t))&|[a,b]&- ∫[a,b] f(t) df(t)所以2∫[a,b] f(t) df(t) = (f(b))^2 - (f(a))^2,所以左边=[(f(b))^2 - (f(a))^2] / 2令x^2+1=2得:x=±1,令x^2+x=2得:x=1或者x=-2注意到x^2+x不总是大于0,令x^2+x=0,得x=0或者x=-1所以S=∫[-2,-1] (x^2+x) dx + ∫[0,1] (x^2+x) dx&- ∫[-1,0] (x^2+x) dx&- (2 * 2 - ∫[-1,1] (x^2+1) dx)=1/2
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1、是的,不过判断函数连续也很简单,初等函数在它们的定义域内都是连续的。初等函数的初等组合(复合,多项式等)也是连续的。3、<img class="ikqb_img" src="http://d./zhidao/wh%3D600%2C800/sign=175d55c6a1cc7cd9fa783cdf09...
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问:当x趋于0如何证明sin(x)/x的极限为1
可以用夹逼定理来证明
1.以(0,0)为圆心,画一个半径为1的圆;
2.作图如下,DA⊥OB,CB⊥OD,直线OC角度为x
3.设三角形ODA面积为S1,扇形面积ODB面积为S2,三角形OCB面积为S3
得S1&=S2&=S3
∵sin(x)=DA/1
∴DA=sin(x)
∵tan(x)=DA/OA
∴OA=cos(x)
∵tan(x)=CB/1
∴CB=tan(x)
∴S1=OA*DA/2=cos(x)*sinx(x)/2
& & S2=π(r^2)*(x/2π)=x/2
& & S3=OB*CB/2=tan(x)/2
& & cos(x)*sin(x)/2&=x/2&=tan(x)/2
化简每项都乘以2/sin(x) 得:
& & cos(x)&=x/sin(x)&=1/cos(x)
& & 1/cos(x)&=sin(x)/x&=cos(x)
∵当x趋近于0时,lim 1/cos(x)=lim cos(x)=1,且sin(x)/x为
∴当x趋近于0时,lim sin(x)/x=1
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