jimmyly said:
Hello HallsoIvy,
Thank you for your response. I just receive quick calculus by taylor and it has what you are explaining at the end of your post. I don't quite understand it at the moment I just started self-studying about a week and a half ago. Can you suggest some resources that may help me understand?
I know of khan academy and mit ocw.
EDIT: I understand what you are explaining up until "typically, what we do is prove the..."
I assume that you have seen the basic
definition of "limit":
\lim_{x\to a} f(x)= L if and only if, given \epsilon> 0, there exist \delta> 0 such that is |x- a|< \delta then |f(x)- L|<\epsilon.
In particular, if f(x)= C, a constant, then |f(x)- C|= |C- C|= 0 which is
always less than any positive \epsilon so \lim_{x\to a} C= C. If f(x)= x, then |f(x)- a|= |x- a|< \epsilon so it is enough to take \delta= \epsilon.
If \lim_{x\to a} f(x)= L, it follows, as above, that given \epsilon> 0, there exist \delta> 0 such that if [/itex]|x- a|< \delta[/itex] then |f(x)- L|< \epsilon. But in that case, |C||f(x)- L|= |Cf(x)- CL|< \epsilon[/itex] also, so
\lim_{x\to a} Cf(x)= CL.
If \lim_{x\to a} f(x)= F and \lim_{x\to a} g(x)= G, then, given \epsilon> 0 there exist \delta_1> 0 such that if |x- a|< \delta_1 then |f(x)- F|< \epsilon/2 and \delta_2> 0 such that if |x- a|< \delta_2 then |g(x)- G|< \epsilon.
(\delta_1 and \delta_2 are not necessarily the same- we will handle that in a minute. Also note the "\epsilon/2". Since \epsilon could be any positive number, and \epsilon/2 is also positive, we can use that as well. You will see why we want \epsilon/2.)
If we define \delta to be the
smaller of \delta_1 and \delta_2, then if |x- a|< \delta we have
both |x- a|< \delta_1 and |x- a|< \delta_2, so that |f(x)+ g(x)- (F- G)|= |f(x)- F+ g(x)- G|\le |f(x)- F|+ |g(x)- G|<\epsilon/2+ \epsilon/2= \epsilon. That is, if \lim_{x\to a} f(x)= F, \lim_{x\to a} g(x)= G, then \lim_{x\to a} f(x)+ g(x)= F+ G.<br />
<br />
Those proofs are in any Calculus text.
