AOL Addition for Functions

maths ✒ aol-functions-addition

Suppose:

Claim

$ (f+g)(x) \to l+m $ as $ x \to a $

Proof

Let $\epsilon > 0 $ be given.

Since $ f(x) \to l ~ $ as $ ~ x \to a $,
$ \exists \delta_1 > 0 $ s.t. $ \forall 0 < |x - a| < \delta_1 $, $ | f(x) - l | < \frac \epsilon 2 $.

Since $ g(x) \to m ~ $ as $ ~ x \to a $,
$ \exists \delta_2 > 0 $ s.t. $ \forall 0 < |x - a| < \delta_2 $, $ | g(x) - m | < \frac \epsilon 2 $.

Write $ \delta = min(\delta_1, \delta_2) > 0 $.

By the Triangle Inequality, $ | (f(x) + g(x)) - (l + m) | ~ \le ~ | f(x) - l | + | g(x) - m | $

Therefore $\forall 0 ~ < ~ |x - a| ~ < ~ \delta $
$ | (f(x) + g(x)) - (l + m) | ~ \le ~ | f(x) - l | + | g(x) - m | ~ < ~ {\frac \epsilon 2}+ {\frac \epsilon 2} ~ = ~ \epsilon $

And so $ (f+g)(x) \to l+m ~ $ as $ ~ x \to a $

$ \square $

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