How to prove? (limit)

**antero** · November 22nd 2009, 08:18 AM

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?

**skeeter** · November 22nd 2009, 08:49 AM

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

**antero** · November 22nd 2009, 09:53 AM

Thank you. But I didn't understand how to find out the value of delta?

**HallsofIvy** · November 22nd 2009, 10:12 AM

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.

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