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.


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