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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?

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    $\begingroup$ 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 $\infty$. $\endgroup$
    – peek-a-boo
    Apr 3 '20 at 16:53
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    $\begingroup$ 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. $\endgroup$
    – Abhi
    Apr 3 '20 at 16:55
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    $\begingroup$ 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 $\infty$). $\endgroup$
    – peek-a-boo
    Apr 3 '20 at 16:58
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    $\begingroup$ @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”. $\endgroup$
    – AJY
    Apr 3 '20 at 17:03
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    $\begingroup$ @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) $\endgroup$
    – peek-a-boo
    Apr 3 '20 at 17:06
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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 $x_0$.

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    $\begingroup$ 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 $\endgroup$
    – Abhi
    Apr 3 '20 at 17:14
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    $\begingroup$ 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. $\endgroup$
    – AJY
    Apr 3 '20 at 17:17
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    $\begingroup$ 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 $\endgroup$
    – Abhi
    Apr 3 '20 at 17:19

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