if $\lim_{x \to c} f(x)$ is exists and $\lim_{x \to c} g(x)$ does not exists, then $\lim_{x \to c} (f(x) + g(x))$ doesn't exists

if $\lim_{x \to c} f(x)$ is exists and $\lim_{x \to c} g(x)$ does not exists, then $\lim_{x \to c} (f(x) + g(x))$ doesn't exists

Proof;

If we show that if $\lim_{x \to c} f(x)$ and $\lim_{x \to c} (f(x) + g(x))$ exists, then $\lim_{x \to c} fg(x)$ exists, we can prove the theorem.

So let assume $\lim_{x \to c} f(x)$ and $\lim_{x \to c} (f(x) + g(x))$ exists, then $$(\forall \epsilon_1 > 0)(\exists \delta_1 > 0 : 0 < |x-c| \leq \delta_1 \rightarrow |f(x)-L_1| < \epsilon_1)$$and$$(\forall \epsilon_2 > 0)(\exists \delta_2 > 0 : 0 < |x-c| \leq \delta_2 \rightarrow |f(x)+g(x)-L_2| < \epsilon_2)$$

So if we choose $\delta_3 \leq min(\delta_1, \delta_2)$, the above conditions will be satisfied for some $\epsilon_1$ and $\epsilon_2$, respectively.In other words,$$-\epsilon_1 < L_1 -f(x) < \epsilon_1$$$$-\epsilon_2 < f(x) + g(x) - L_2 < \epsilon_2$$

If we add 2 inequality and choose $\epsilon_3=\epsilon_1 + \epsilon_2$ and $L_3 = L_2 - L_1$, we get

$$|g(x)-L_3| < \epsilon_3$$, which is satisfied for all $\delta_3> 0$, so the theorem is proved.

Is there any mistake in the proof ? or do you have any advice about the construction of the proof or how I expressed while giving the proof ?