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)

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}[