# Prove that $\infty+c$ is not an indeterminate form, where $c$ is a real number

So, here's what I'm trying to prove:

Let $$f$$ and $$g$$ be functions. Suppose that $$f(x) \to c$$ as $$x \to x_0$$ and $$g(x) \to \infty$$ as $$x \to x_0$$. Suppose that $$c$$ is finite. Then, $$f(x)+g(x) \to \infty$$ as $$x \to x_0$$.

Proof Attempt:

Let $$M>0$$. We have to prove that there is a $$\delta>0$$ such that:

$$0 < |x-x_0| < \delta \implies |f(x)+g(x)| > M$$

Now, we know that:

$$|f(x)+g(x)| \geq |g(x)|-|f(x)|$$

and we want that right-hand side to be greater than $$M$$. In other words, $$|g(x)| > |f(x)| + M$$. Let $$\epsilon > 0$$. Then, there exists a $$\delta_1 > 0$$ such that:

$$0 < |x-x_0| < \delta_1 \implies |f(x)-c| < \epsilon$$

$$\implies |f(x)| < |c| + \epsilon$$

Now, let $$\delta_2 > 0$$ exist such that:

0 < |x-x_0| < \delta_2 \implies |g(x)| > M + |c| + \epsilon$Define $$\delta = \min\{\delta_1,\delta_2\}$$. Then: $$0 < |x-x_0| < \delta \implies |f(x)+g(x)| \geq |g(x)| - |f(x)| > M + |c| + \epsilon - |c| - \epsilon = M$$ Since the existence of the desired $$\delta$$ has been established, this proves the desired result. Does the proof above work? If not, why? How could I, then, improve it? • actually, since you're trying to prove$f(x) + g(x) \to \infty$as$x \to x_0$, you have to prove that$0< |x-x_0|< \delta \implies f(x)+g(x) > M$(without the absolute values). This is because you want to prove the function itself goes to$\infty$, not its absolute value apporahes$\infty$. Apr 3 '20 at 16:53 • Uh the definition given in my text is as follows:$lim_{x \to x_0} f(x) = \infty \iff \forall M>0: \exists \delta > 0: 0 < |x-x_0| < \delta \implies |f(x)| > M$. Like, that's what I'm working with so I assume that it applies for this problem. – Abhi Apr 3 '20 at 16:55 • are you sure? because that definition is not the correct one. According to that definition, the function$f(x) = -\dfrac{1}{x^2}$would satisfy$\lim_{x \to 0}f(x) = \infty$, when in fact the limit should be$-\infty$. It would also assign a limit of$\infty$to functions whose limit doesn't exist (for example consider$e^x \sin x$as$x \to \infty$. This has no limit, because changes sign repeatedly, but its absolute value does have a limit$\infty$). Apr 3 '20 at 16:58 • @peek-a-boo In some contexts, we use this definition for a limit of$\infty$. Sometimes we don’t need to think about$\infty$as being signed, but instead as just “really big”. – AJY Apr 3 '20 at 17:03 • @AJY I see. If that's the case, then ok... it's just not a definition I've seen in any book (Spivak,Rudin, Duistermaat). And about the exponential thing yes of course, you're right; clearly I have messed up what I intended to say (and at this point I don't even remember what I intended to say lol) Apr 3 '20 at 17:06 ## 1 Answer This looks correct! One small thing perhaps worth noting is that you didn’t have to do this for all $$\epsilon > 0$$, but could’ve just chosen $$\delta > 0$$ such that $$0<|x - x_0| < \delta_1 \Rightarrow |f(x) - c| < 7, |g(x)| > M + |c| + 7$$, or any other positive number. In fact, $$f$$ doesn’t even need to have a limit at $$x_0$$, it just needs to bounded in some neighborhood of $$x_0$$. • Oh nice! I will try proving it for a function that is bounded in some neighbourhood of$x_0$. I probably won't remember that theorem specifically, though, cos I suck at remembering things. I'm, like, working straight off of the definition most of the time – Abhi Apr 3 '20 at 17:14 • I don’t know if it’s a terribly important thing to know in itself, it’s just good sometimes once you’ve written a proof to take stock of what assumptions you used in the process. The general idea in this is that if$g$is going to$\infty$, and$f\$ is only so big around there, then infinite plus bounded is still infinite.
– AJY
Apr 3 '20 at 17:17
• Yeap. The general ideas are easy to grasp. They're intuitively obvious, even. I just need to revisit the assumptions being made so that I'm not being stupid when applying a theorem in solving a problem. I've mostly been covering up the proofs given in the book and have been trying to prove the results on my own. So, I sort of only pay attention to the assumptions when I need to use them in proving the result
– Abhi
Apr 3 '20 at 17:19