一道微积分题f(x) = e^(-1/x²) if x 不等于0,f(x) = 0 if x = 0.a.us

不可以用泰勒公式,因为泰勒公式假定极限存在,而题目是要我们证明极限存在
a.use the definition of derivative to show that f is differentiable at x = 0.What is f '(0)?
利用导数的定义证明f在x=0处可导.f'(0)=?
f(x) 在x=0处可导 当且仅当 下面两个极限存在并相等：
lim h->0+ (f(h)-f(0))/h = lim h->0- (f(h)-f(0))/h
lim h->0+ (f(h)-f(0))/h
= lim h->0+ e^(-1/h²)/h
= lim h->+∞ e^(-h²)*h
= lim h->+∞ h/e^(h²) 接着用洛必达法则
= lim h->+∞ 1/(2h e^(h²))
= 0
lim h->0- (f(h)-f(0))/h
= lim h->0- e^(-1/h²)/h
= lim h->-∞ e^(-h²)*h
= lim h->-∞ h/e^(h²) 接着用洛必达法则
= lim h->-∞ 1/(2h e^(h²))
= 0
故f(x) 在x=0处可导,f'(0) = 0
b.show that lim x->0 [e^(-1/x²)/x^n] = 0 for every positive integer n
证明对任意正整数n,lim x->0 [e^(-1/x²)/x^n] = 0
证：
lim x->0 [e^(-1/x²)/x^n]
= lim x->∞ [e^(-x²)*x^n]
= lim x->∞ [x^n/e^(x²)] 接着用洛必达法则
= lim x->∞ [nx^(n-1)/(2x*e^(x²))]
= lim x->∞ [nx^(n-2)/(2*e^(x²))] 接着重复使用洛必达法则直至x的指数小于等于0为止
= ...
=
lim x->∞ [n(n-2)(n-4)...2/(2^(n/2)*e^(x²))] (n 为偶数)
lim x->∞ [n(n-2)(n-4)...1/(2^((n+1)/2)x*e^(x²))] (n 为奇数)
= 0

对于x=0点的导数，必须用定义式来求解。
根据定义， f '(0)=limh→0{[f(0+h)-f(0)]/h}，因为x=0时，f(0)=0,所以，
即f '(0)=limh→0[f(h)/h]
f(h)=e^(-1/h²)=1/[e^(1/h²)]
则f(h)/h=(1/h)/[e^(1/h²)]
令1/h=t则f...

只写关键的部分啊。
所用条件有 h->0；当x->0时，e^x=x+1。
[e^(-1/(x+h)²)-e^(-1/x²)]/h=e^(-1/x²)[e^((1/x²)-(1/(x+h)²))-1]/h=e^(-1/x²)[(1/x²)-(1/(x+h)²)+1-1]=e^(-1/x²)*（h...