Prove that the sum of two continuous functions is continuous using epsilon delta.

Question

$$\lim _{x\rightarrow a} f(x)+g(x) = f(a)+g(a)$$

Let $ε>0$ be given

Since $f$ and $g$ are continuous, $|f(x)-f(a)|< ε$ when $0<|x-a|< \delta_f $ and $|g(x)-g(a)|< ε$ $\ $ when $0<|x-a|< _g$

Let $_h$ be defined as $\min(_g ,_f)$ and $h(x)$ be defined as $f(x)+g(x)$

$\ |f(x)-f(a)|+|g(x)-g(a)|< 2ε$

$\ |f(x)-f(a) + g(x)-g(a)|< 2ε$

$\ |h(x)-h(a)|< 2ε$

We can replace $_f$ and $_g$ by $_h$ to get

$|f(x)-f(a)| < ε$ when $0<|x-a|<_h$ and $\ |g(x)-g(a)|< ε$ when $0<|x-a|<_h$

How do I prove that $\ |h(x)-h(a)|< ε$ instead of $\ |h(x)-h(a)|< 2ε$?

Answer 1

Start your proof as follows: Let $\varepsilon >0$ be given. Then there are $\delta_f>0$ and $ \delta_g>0$ with

$|f(x)-f(a)|< ε/2$ when $|x-a|< \delta_f$ and $|g(x)-g(a)|< ε/2$ when $|x-a|< \delta_g$....

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