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The proposition starts with suppose that the limit of $f(x)=a$ as $x\to{p}$ and the limit of $g(x)=b$ as $x\to{p}$.

I know we have to start of with the definition of what the prop starts of with. But then I don't know how to proceed from there.

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1 Answer 1

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Given. We have that for all $\epsilon > 0$, there exists a $\delta > 0$ such that $\delta > |x-p| > 0$ implies:

$$ |f(x) - a| < \epsilon/2$$ $$ |g(x) - b| < \epsilon/2$$

Rest of the proof. Adding these two inequalities,

$$|f(x) - a| + |g(x) - b| < \epsilon.$$

By the Triangle Inequality, the LHS is greater than $|(f(x)-a) + (g(x) - b)| = |(f(x)+g(x)) - (a+b)|$, and so

$$|(f(x)+g(x)) - (a+b)| < \epsilon.$$

So, for all $\epsilon > 0$, there exists an $\delta > 0$ such that $\delta > |x-p| > 0$ implies

$$|(f(x)+g(x)) - (a+b)| < \epsilon.$$

Thus the limit as $x \rightarrow p$ of $(f+g)(x) = a+b$.

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Awesome I got how u used the triangle inequality! I understood it thank you! –  Maximiliano Oct 15 '12 at 19:50
    
Sure thing. We have (by the Triangle Inequality): $$|f(x) - a| + |g(x) - b| \geq |f(x) - a + g(x) - b| = |(f+g)(x) - (a+b)|.$$ Since $\epsilon > |f(x) - a| + |g(x) - b|$, we also have $\epsilon > |(f+g)(x) - (a+b)|$. –  Michael Zhao Oct 15 '12 at 19:50

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