求导,求极限

题目

求导,求极限
1.已知f(x)= x^2sin(1/x) x不等于0时, f(x)=0 x=0时. 求f'(x)
2.求极限：lim(x->0)[(e^x-1-x)^2/tan(sinx)^2]

答案

1.已知f(x)= x^2sin(1/x) x不等于0时,f(x)=0 x=0时.求f'(x)
x不等于0时,f'(x) = 2xsin(1/x) + x^2cos(1/x)*(-1/x^2)
= 2xsin(1/x) - cos(1/x)
x = 0 时,
lim_{x->0} [f(x)] = lim_{x->0}[x^2sin(1/x)] = 0 = f(0),
所以,f(x)在 x = 0处连续.
lim_{x->0}{[f(x) - f(0)]/(x-0)} = lim_{x->0}[x^2sin(1/x)/x]
= lim_{x->0}[xsin(1/x)] = 0
所以,f(x)在 x = 0处可导,f'(0) = 0.
综合,有,
x不等于0时,f'(x) = 2xsin(1/x) - cos(1/x)
x = 0 时,f'(0) = 0.
2.求极限：lim(x->0)[(e^x-1-x)^2/tan(sinx)^2]
lim_{x->0}[(e^x-1-x)^2/tan(sinx)^2]
= lim_{x->0}[(e^x-1-x)^2/(sinx)^2][(sinx)^2/tan(sinx)^2]
= lim_{x->0}[(e^x-1-x)^2/(sinx)^2]
= lim_{x->0}[(e^x-1-x)^2/x^2] [x^2/(sinx)^2]
= lim_{x->0}[(e^x-1-x)^2/x^2]
= lim_{x->0}[2(e^x-1-x)(e^x - 1)/(2x)]
= lim_{x->0}[e^x-1-x][(e^x - 1)/x]
= lim_{x->0}[e^x-1-x]
= 1 - 1 - 0
= 0.