3. Let $p$ be a real number and $c \neq 0$ an integer such that
$$
c-0.1<x^{p}\left(\frac{1-(1+x)^{10}}{1+(1+x)^{10}}\right)<c+0.1
$$
for all (positive) real numbers $x$ with $0<x<10^{-100}$. (The exact value $10^{-100}$ is not important. You could replace it with any "sufficiently small number".)
Find the ordered pair $(p, c)$.
${ }^{1}$ Note that this does actually have integer solutions by Bezout's identity, as $\operatorname{gcd}(13 \cdot 31,5 \cdot 31,5 \cdot 13)=1$.
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 $f(x):=\frac{1-(1+x)^{10}}{1+(1+x)^{10}}=\frac{-10 x+O\left(x^{2}\right)}{2+O(x)}$, where the "big-O" notation simply make precise the notion of "error terms". ${ }^{2}$
Intuitively, $f(x) \approx \frac{-10 x}{2}=-5 x$ 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 ).

Remark. Why does $(p, c)=(-1,-5)$ actually satisfy the inequality? This is where the $10^{-100}$ kicks in: for such small values of $x$, the "error" $|g(x)-(-5)|$ of the approximation $g \approx-5$ does actually lie within the permitted threshold of $\pm 0.1$. (You can easily work out the details yourself, if you're interested. It's something you might want to work out once or twice in your life, but rational functions are "well-behaved" enough that we can usually rely on our intuition in these kinds of scenarios.)