Recent content by andilus

Sorry,I have not express clearly. what i want to prove is just: lim(x->a)(\frac{f(x)-f(a)}{x-a})=lim(h->0)(\frac{f(a+h)-f(a)}{h})

how to prove lim(x->a)\frac{f(x)-f(a)}{x-a}=lim(h->0)\frac{f(a+h)-f(a)}{h} it seems to be obvious, but i don't know how to prove```

i seem to know... if x0<a, obviously wrong. if x0>b, we can find some x such that b-\delta<x<b satisfying that x is upper bound of the set.

but how to prove x0 is in [a,b]?

what does "the set of rational numbers" mean? Is {1,2,3} a set of rational numbers?

According to supremum and infimum principle, nonempty set A={x|x\inQ,x2<2} is upper bounded, so it should have a least upper bound. In fact, it dose not have least upper bound. Why? When the principle is valid?

Homework Statement sup problem if f is continuous on [a,b] with f(a)<0<f(b), show that there is a largest x in [a,b] with f(x)=0Homework Equationsi think it can be done by least upper bounds, but i dun know wat is the exact prove. The Attempt at a Solution

if f is continuous on [a,b] with f(a)<0<f(b), show that there is a largest x in [a,b] with f(x)=0 i think it can be done by least upper bounds, but i dun know wat is the exact prove.