Problem 1639

Solution:
Answer: $(-1,-5)$ This is essentially a problem about limits, but phrased concretely in terms of "small numbers" (like 0.1 and $10^{-100}$).

We are essentially studying the rational function $f(x) := \frac{1-(1+x)^{10}}{1+(1+x)^{10}} = \frac{-10x + O(x^{2})}{2 + O(x)}$, where the "big-O" notation simply makes precise the notion of "error terms".

Intuitively, $f(x) \approx \frac{-10x}{2} = -5x$ for "small nonzero $x$". (We could easily make this more precise if we wanted to, by specifying the error terms more carefully, but it's not so important.) So $g(x) := x^{p} f(x) \approx -5 x^{p+1}$ for "small nonzero $x$".

- If $p+1 > 0$, $g$ will approach $0$ ("get very small") as $x$ approaches $0$ (often denoted $x \rightarrow 0$), so there's no way it can stay above the lower bound $c-0.1$ for all small nonzero $x$.

- If $p+1 < 0$, $g$ will approach $-\infty$ ("get very large in the negative direction") as $x \rightarrow 0$, so there's no way it can stay below the upper bound $c+0.1$ for all small nonzero $x$.

- If $p+1 = 0$, $g \approx -5$ becomes approximately constant as $x \rightarrow 0$. Since $c$ is an integer, we must have $c = -5$ (as $-5$ is the only integer within $0.1$ of $-5$).