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点x=A是函数Fx=(x%A)sin1/(x%A)的( )间断点

(2) lim(x->a) (x-a) sin(1/(x-a)) =0 ans : A (3) f(x) =|x-1| lim(x->1+) f(x) = 0 lim(x->1-) f(x) = 0 x=0, f(x)连续 f'(1+) =lim(h->0+) [f(1+h) - f(1)] /h =lim(h->0) h/h =1 f'(1-) =lim(h->0) [f(1-h) - f(1)] /h =lim(h->0) -h/h =-1 ...

待续

lim x->0 x^a sin(1/x)=f(0) sin(无穷)属于【0,1】 所以 limx->0 x^a=0 a>0 f(x)连续 2) limx->0(x^a sin(1/x))’=f'(0) x^a sin(1/x)'=ax^(a-1)sin(1/x)-cos(1/x)*(x)^(-2)x^a =ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) f'(0)=lim h->0 [f(h)-f(0)]/...

由题意知,f(x)在x=0处连续,所以:f(0)=b=limx→0?x2sin1x=0 (由于.sin1x.≤1)同时,f(x)在0点左右导数相等,limx→0?x2sin1x?f(0)x?0=a即:limx→0?xsin1x=0所以a=b=0故a2+b2=0

供参考。

f(x) =0 ; x≤0 =x^a.sin(1/x) ; x>0 f(0)=f(0-)=lim(x->0) 0 =0 f(0+) =lim(x->0) x^a. sin(1/x) =0 =>a>0 x=0, f(x) 连续,a>0 f'(0+)=0 f(0-) =lim(h->0) [ h^a. sin(1/h) - f(0) ]/h =lim(h->0) h^(a-1). sin(1/h) f'(0) 存在 => a-1 >0 a>1 ...

f(x)=x^a·sin(1/x) f'(x)=ax^(a-1)sin(1/x)-x^a·cos(1/x)/x² =x^(a-2)[axsin(1/x)-cos(1/x)] a>2时 lim(x→0)f'(x)=0 f'(x)连续 a=2 f'(x)=2xsin(1/x)-cos(1/x) lim(x→0)f'(x)=0-cos(1/x) ∵cosx是偶函数 ∴左极限=右极限=函数值 f'(x)连续 a

 是x趋于无穷g(x)=(1+1/x)^x 的极限是e所以令a=1/x则a趋于无穷所以(1+x)^(1/x)=(1+1/a)^a所以极限是e

原题是:f(x)=x-(1/3)sin2x+asinx在(-∞,+∞)上递增,求a的取值范围. f'(x)=1-(2/3)cos2x+acosx =1-(2/3)(2cos²x-1)+acosx =-(4/3)cos²x+acosx+(5/3) 设t=cosx f'(x)=g(t)=-(4/3)t²+at+(5/3),-1≤t≤1 g(t)=-(4/3)t²+at+(5/3)是...

由f(x)的表达式,可以看出只有x=0,1,2是它的不连续点(1)选项A,limx→?1+f(x)=sin(?1?2)?1?(?2)?(?1?2)2=?sin318limx→0?f(x)=limx→0??xsin(x?2)x(x?1)(x?2)2=?sin24因此,f(x)在x=-1处有右极限,在x=0处有左极限,而f(x)在(-1,0...

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