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The question is to use two functions f(x) and g(x) to show that ∞ - ∞ (infinity - infinity) is indeterminate.
I don't really know how to get started on this. I think I need to find two limits for f(x) and g(x) that are both infinity. But how to I show that subtracting them is indeterminate?

Edit: In the example we were shown, lim x --> 0 was used. But that was for the indeterminate form 0∞. I don't know if that applies here or not.

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"Indeterminate" means "not having a unique value" as $x$ tends to whatever the problem says (I am guessing $x \rightarrow 0$ or $x \rightarrow \infty$, but you really should specify and not make the community you are asking for help guess). I will assume $x \rightarrow \infty$.

So, to show indeterminacy, I will exhibit three examples of the function pair $f(x), g(x)$ with different limiting behaviors of the difference $[f(x)-g(x)]$ as $x \rightarrow \infty$.

Example 1: $f(x) = g(x) = x$. Here $\lim_{x \rightarrow \infty}[f(x) - g(x)] = 0$.

Example 2: $f(x) = x^2, \; g(x) = x$.

Example 3: See Mark Fischler's answer.:)

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  • $\begingroup$ Thank you for the reply. The part that I don't understand is that when you use different values for f and g. Wouldn't you expect a different answer? I don't really understand how that makes it indeterminate. $\endgroup$ – user372991 Sep 28 '16 at 21:58
  • $\begingroup$ Well consider a form that is not indeterminate: $1 / \infty$. In detail: if you have $$ \lim_{x \rightarrow \infty} f(x) = \mbox{(some real cosntant)}, \quad \lim_{x \rightarrow \infty} g(x) = \infty, $$ then, no matter which $f, g$ you use, as long as they satisfy the above conditions, you will get $$ \lim_{x \rightarrow} {f(x) \over g(x)} = 0. $$ In a high-level summary: The goal is, knowing the asymptotic behavior of $f, g$, to be able to ascertain the asymptotic behavior of an expression involving these two. When we cannot do that, the latter expression is indeterminate. $\endgroup$ – user8960 Sep 28 '16 at 23:18
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Consider $$ f(x) = \csc^2x \\ g(x) = \cot^2 x \\ h(x) = 1/x^2 $$ and look at the limit as $x\to 0$ of the expressions $f(x)-g(x)$ and $f(x)-h(x)$.

Both of these expressions are of the form $\infty - \infty$

But $$ \lim_{x\to 0} (f(x) - g(x)) = 1 \\ \lim_{x\to 0} (f(x) - h(x)) = \frac13 $$ This shows that $\infty - \infty$ could be $1$ or $\frac13$; for that matter it could be $0$ (consider $f(x)-f(x)$).

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  • $\begingroup$ If you wouldn't mind, could you explain how f(x) -g(x) = 1 and f(x) - h(x) = 1/3? I thought that they all would equal infinity? $\endgroup$ – user372991 Sep 28 '16 at 22:18
  • $\begingroup$ @Jordan He is just giving you the answers for those limits. You can verify them with L'Hospital's Rule. It's a good exercise $\endgroup$ – imranfat Sep 28 '16 at 22:52
  • $\begingroup$ On the other hand, the first one is particularly easy: Since $csc^2x - \cot^2x = 1$ for all values of angles where they are both finite, "obviously" the limit as $x\to$ any real value (or for that matter even any complex value) must be $1$. $\endgroup$ – Mark Fischler Oct 1 '16 at 21:20

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