高手对于英语也无能为力啊

About the first question
Prove：think of   |x|<1 ε>0,
then
if x is rational, |f(x)|= |x|2<|x|<ε; if x is irrational, obviously,|f(x)|=0<ε.
last so we could get this
δ=ε, while |x|<δ, |f(x)|<ε, as this, lim x→0f(x)=0 is proved
i"m such a lazy guy
so
let someone else help you solve the nxet problem  


原谅我每层楼只能插一张图片  
这种类似的题目其实还可以就是你是刚刚接触吧
这是影印版微积分上的嘛 的确有点麻烦的还有不懂的你可以加我QQ问 这里说麻烦
补充刚刚的
last ，so we could get this
δ=ε, while |x|<δ, |f(x)|<ε, as this, lim x→0f(x)=0 is proved


好厉害好厉害我爱死你了 谢谢谢谢谢谢

回复：4楼
是影印版微积分！！！！！太感谢你了 能给我一下你QQ吗，我拜你为师！！！

Let p(x)=(c_n)x^n+...(c_1)x+c_0 be a polynomial.
Since lim_{x→a}f(x)g(x)=lim_{x→a}f(x)*lim_{x→a}g(x) as long as both the two limits at the right side exist, we have lim_{x→a}(c_k)x^k=(lim_{x→a}c_k)(lim_{x→a}x)^k=(c_k)a^k for any positive integer k.
Moreover, since lim_{x→a}(f(x)＋g(x))=lim_{x→a}f(x)+lim_{x→a}g(x) as long as both the two limits at the right side exist, we have lim_{x→a}p(x)=∑_{k=0 to n}lim_{x→a}(c_k)x^k=∑_{k=0 to n}(c_k)a^k=p(a)


回复：6楼
你膜拜8楼的就好 人家才是大湿

奇怪7楼呢？

回复：8楼
拜你为师吧！！！！！！！！ 谢谢谢谢

回复：9楼
谢谢！！能否给我一个QQ？

回复：10楼
我也想知道

回复：12楼
想了想我还是不能误了你孩子向前看齐吧

同14楼

都太了不起了