How to prove? (limit)

• Nov 22nd 2009, 08:18 AM
antero
How to prove? (limit)
How to prove lim(f(x)+g(x))= a+b as x->x0 and lim f(x)=a, lim g(x)=b? Could someone help me?
• Nov 22nd 2009, 08:49 AM
skeeter
Quote:

Originally Posted by antero
How to prove lim(f(x)+g(x))= a+b as x->x0 and lim f(x)=a, lim g(x)=b? Could someone help me?

proofs of various limit properties here ...

Pauls Online Notes : Calculus I - Proof of Various Limit Properties
• Nov 22nd 2009, 09:53 AM
antero
Thank you. But I didn't understand how to find out the value of delta?
• Nov 22nd 2009, 10:12 AM
HallsofIvy
Look at |f(x)+g(x)- (a+b)|= |(f(x)-a)+(g(x)-b)| and use the fact that $|(f(x)-a)+ (g(x)-b)|\le |f(x)-a|+ |g(x)- b|$. In order to make $|(f(x)-a)+ (g(x)-b)|< \epsilon$, we have to make sure that |f(x)-a| and |g(x)-b| add up to something less than $\epsilon$. One way to do that is to make sure that |f(x)-a| and |g(x)-b| each are less than $\epsilon/2$. Since $\lim_{x\to x_0} f(x)= a$, there exist some $\delta_1$ such that if [tex]|x- x_0|< \delta_1[\math], $|f(x)- a|< \epsilon/2$. Since $\lim_{x\to x_0} g(x)= b$, there exist some $\delta_2$ such that if $|x- x_0|< \delta_2$, $|g(x)- b|< \epsilon/2$.

To be sure that both of those happen, choose $\delta$ to be less than or equal to the smaller of $\delta_1$ and $\delta_2$. That way, if $|x-x_0|< \delta$, both $|x-x_0|< \delta_1$ and $|x-x_0|< \delta_2$ are true.