0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.