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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

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 ...

实际上就是求cos(1/x),sin(1/x)的阶,而这两个当x->0时极限是不存在的,但是有界,所以只要a>0即可保证极限为0。

不连续,x趋近于0正(Sin1/x是有界函数乘以0等于0)是等于0的,x趋近于0负是等于1(sinx和x是等价无穷小)的

lim x->0- 1/x·sin2x=a=lim x->0+ x·sin1/x+b2=a=0+b  (利用lim x->0 sin2x/x=2,  lim x->0 xsin1/x=0)所以a=2,b=2

f ' (0) = Limit [ ( f(x)-f(0) ) / (x-0) , x->0 ] = Limit [ x^2 sin(1/x) / x , x->0 ] = Limit [ x sin(1/x) , x->0 ] = 0 选 D 注:sin(1/x) 有界,无穷小与有界函数的乘积仍是无穷小, Limit [ x sin(1/x) , x->0 ] = 0

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...

原极限=lim(x趋于0) (x^2 -sin^2x) / (x^2*sin^2x) =lim(x趋于0) (x -sinx)(x+sinx) / (x^2*sin^2x) x趋于0的时候,sinx等价于x 所以得到x+sinx等价于2x, x^2*sin^2x等价于x^4 所以原极限=lim(x趋于0) 2x *(x -sinx) / x^4 =lim(x趋于0) 2(x -s...

f(x)=|x|sin(x-2)/x(x-1)(x-2) lim(x→0-)f(x)=lim(x→0-)-sin(x-2)/(x-1)(x-2)=sin(2)/2→A区间内有界 lim(x→1-)f(x)=lim(x→1-)sin(x-2)/(x-1)(x-2)=-∞ lim(x→1+)f(x)=lim(x→1+)sin(x-2)/(x-1)(x-2)=+∞→BC区间内无界 lim(x→2+)f(x)=lim(x→2+)sin(x...

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