Question

Using the epsilon-delta definition of the limit prove that if lim x-->a f(x) and lim x-->a g(x) exist, then lim x-->a [f(x)+g(x)] = lim x-->a f(x) + lim x-->a g(x).

Answer 1

Conditions

Using the epsilon-delta definition of the limit, prove that if $\lim x \to f(x)$ and $\lim x \to g(x)$ exist, then $\lim x \to f(x) + g(x) = \lim x \to f(x) + \lim x \to g(x)$.

Solution

Definition. The limit of function $f(x)$ is equal to $F$ (when $x \to a$), if:

$$\forall \varepsilon > 0 \ \exists \delta = \delta (\varepsilon) \ \forall x: |x - \alpha| < \delta \ |f(x) - F| < \varepsilon$$

Let's write, what means that $f(x)$ and $g(x)$ have limits when $x \to a$:

$$\forall \varepsilon > 0 \ \exists \delta_1 = \delta_1 (\varepsilon) \ \forall x: |x - \alpha| < \delta_1 \ |f(x) - F| < \varepsilon$$ $$\forall \varepsilon > 0 \ \exists \delta_2 = \delta_2 (\varepsilon) \ \forall x: |x - \alpha| < \delta_2 \ |g(x) - G| < \varepsilon$$

Fix $\varepsilon > 0$, consider $\delta = \max(\delta_1; \delta_2)$

$$|f(x) - F + g(x) - G| \leq |f(x) + g(x) - (F + G)| \leq |f(x) - F| + |g(x) - G| < 2\varepsilon$$

As we can see:

$$\forall \varepsilon > 0 \ \exists \delta = \delta (\varepsilon) = \max(\delta_1; \delta_2) \ \forall x: |x - \alpha| < \delta \ |f(x) + g(x) - (F + G)| < 2\varepsilon$$

This means, that $f(x) + g(x)$ has a limit and it exactly equal to the sum of $F$ and $G$.

Q.E.D.