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.
On
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 $f(x)-f(x)$).
"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.:)