# Proof for the existence of a limit using the $(\varepsilon, \delta)$ definition

Question: Suppose that $$\lim_{z \to z_0}g(z)$$ exists and is finite. Assume that $$f$$ is another function defined on the same domain as $$g$$. Show that $$\lim_{z \to z_0}f(z)$$ exists iff $$\lim_{z \to z_0}f(z) +g(z)$$ exists.

My Proof: I believe it's sufficient to state that if $$\lim_{z \to z_0}g(z) = L$$ and $$\lim_{z \to z_0}f(z) +g(z) = K$$, then

\begin{align} \lim_{z \to z_0}f(z) +g(z) &= K\\ \left[\lim_{z \to z_0}f(z) \right] + \left[\lim_{z \to z_0}g(z) \right] &= K\\ \left[\lim_{z \to z_0}f(z) \right] + L &= K\\ \lim_{z \to z_0}f(z) &= K - L. \end{align}

How can I verify this result with the epsilon-delta definition of a limit? My attempt so far is as follows:

Since we are given that $$\lim_{z \to z_0}g(z) = L$$, we know that $$\forall \varepsilon, \exists\delta$$ such that $$|g(z) - L| < \varepsilon$$ and $$0<|z-z_0|<\delta$$. Similarly, we know that $$\lim_{z \to z_0}f(z) +g(z) = K$$, and so $$\forall \varepsilon', \exists\delta'$$ such that $$|f(z) + g(z) - K| < \varepsilon'$$ and $$0<|z-z_0|<\delta'$$. If the limit does exist, then $$\forall \varepsilon'', \exists\delta''$$, we must show that $$|f(z) - (K-L)| < \varepsilon''$$ and $$0<|z-z_0|<\delta''$$.

I believe this setup is correct, however I'm unsure where to go from here. Any help or guidance would be greatly appreciated!

Suppose that $$\lim_{z \to z_0} (f(z)+g(z)) = K$$ and $$\lim_{z \to z_0} g(z) = L$$. Now, we want to show that:

$$\lim_{z \to z_0} f(z) = K-L$$

Let $$\epsilon > 0$$ be given. Then, we want a $$\delta > 0$$ such that:

$$0 < |z-z_0| < \delta \implies |f(z)-(K-L)| < \epsilon$$

Then, we can see that:

$$|f(z)-(K-L)| = |(f(z)+g(z)-K)+(L-g(z))| \leq |(f(z)+g(z))-K| + |L-g(z)|$$

We know that:

$$\exists \delta_1 > 0: 0 < |z-z_0| < \delta_1 \implies |(f(z)+g(z))-K| < \frac{1}{2}\epsilon$$

$$\exists \delta_2 > 0: 0 < |z-z_0| < \delta_2 \implies |g(z)-L| < \frac{1}{2} \epsilon$$

Define $$\delta = \max\{\delta_1,\delta_2\}$$. Then:

$$0 < |z-z_0| < \delta \implies |f(z)-(K-L)| \leq |(f(z)+g(z))-K| + |g(z)-L| < \epsilon$$

which proves the desired result. I will leave you to try and prove the forward conditional on your own. $$\Box$$

I hope that makes sense.