设函数f(x)f(x)f(x)在x=1x = 1x=1处连续,且limx→1f(x)=6\lim\limits_{x \to 1}f(x)=6x→1limf(x)=6,则f(1)=f(1)=f(1)=( )。
(A)1
(B)0
(C)6
(D)−6-6−6
已知函数y=2arctanxy = 2\arctan xy=2arctanx,则y′=y'=y′=( )。
(A)11+x2\dfrac{1}{1 + x^{2}}1+x21
(B)−11+x2-\dfrac{1}{1 + x^{2}}−1+x21
(C)21+x2\dfrac{2}{1 + x^{2}}1+x22
(D)−21+x2-\dfrac{2}{1 + x^{2}}−1+x22
已知x→0x \to 0x→0时,f(x)→0f(x) \to 0f(x)→0,且f(x)≠0f(x) \neq 0f(x)=0,则x→0x \to 0x→0时,1f(x)\dfrac{1}{f(x)}f(x)1是无穷大.( )
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