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?

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- Nov 22nd 2009, 07:18 AManteroHow 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, 07:49 AMskeeter
proofs of various limit properties here ...

Pauls Online Notes : Calculus I - Proof of Various Limit Properties - Nov 22nd 2009, 08:53 AMantero
Thank you. But I didn't understand how to find out the value of delta?

- Nov 22nd 2009, 09:12 AMHallsofIvy
Look at |f(x)+g(x)- (a+b)|= |(f(x)-a)+(g(x)-b)| and use the fact that $\displaystyle |(f(x)-a)+ (g(x)-b)|\le |f(x)-a|+ |g(x)- b|$. In order to make $\displaystyle |(f(x)-a)+ (g(x)-b)|< \epsilon$, we have to make sure that |f(x)-a| and |g(x)-b| add up to something less than $\displaystyle \epsilon$. One way to do that is to make sure that |f(x)-a| and |g(x)-b| each are less than $\displaystyle \epsilon/2$. Since $\displaystyle \lim_{x\to x_0} f(x)= a$, there exist some $\displaystyle \delta_1$ such that if [tex]|x- x_0|< \delta_1[\math], $\displaystyle |f(x)- a|< \epsilon/2$. Since $\displaystyle \lim_{x\to x_0} g(x)= b$, there exist some $\displaystyle \delta_2$ such that if $\displaystyle |x- x_0|< \delta_2$, $\displaystyle |g(x)- b|< \epsilon/2$.

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