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Proof: $\lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$

Let $\lim_{x \to a} f(x) = L_1$ and $\lim_{x \to a} g(x) = L_2$, where

$\forall \epsilon > 0, \exists \delta_1 > 0 : (0 < |x-a| < \delta_1 \implies |f(x) - L_1| < \epsilon / 2)$

$\forall \epsilon > 0, \exists \delta_2 > 0 : (0 < |x-a| < \delta_2 \implies |g(x) - L_2| < \epsilon / 2)$

We wish to show that

$\forall \epsilon > 0, \exists \delta > 0 : (0 < |x-a| < \delta \implies |f(x) + g(x) - (L_1 + L_2)| < \epsilon)$

By the triangle inequality $|f(x) + g(x) - (L_1 + L_2)| \leq |f(x) - L_1| + |g(x) - L_2| < \epsilon$

If we set $\delta = \min(\delta_1, \delta_2)$, then $|f(x) - L_1| + |g(x) - L_2| < \epsilon/2 + \epsilon/2 = \epsilon$.

Is this a correct proof?

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    $\begingroup$ Absolutely. Your argument is fine. $\endgroup$ Mar 7 '18 at 5:29
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All you have to do is to remove $\epsilon$ from:

By the triangle inequality $|f(x)+g(x)−(L1+L2)|≤|f(x)−L1|+|g(x)−L2|<ϵ$

( change to )

By the triangle inequality $|f(x)+g(x)−(L1+L2)|≤|f(x)−L1|+|g(x)−L2|$

If we set $δ=min(δ1,δ2)$, then $|f(x)−L1|+|g(x)−L2|<ϵ/2+ϵ/2=ϵ.$

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  • $\begingroup$ I am confused, what is wrong with the epsilon? Did I make a mistake? $\endgroup$
    – user537069
    Mar 7 '18 at 6:33
  • $\begingroup$ It wasn't wrong, it was misplaced. $\endgroup$ Mar 7 '18 at 8:04

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