The statement \(\lim\limits_{x \rightarrow 0} 1/x^2= \infty\) certainly makes intuitive sense. Construct a rigorous definition in the *challenge-response* style for a limit statement of the form \(\lim\limits_{x \rightarrow c} f(x) = \infty\) and use it to prove the previous statement.
Definitions and Theorems:
- Definition of functional limits
- [4.2.3] Sequential Criterion for Functional Limits
- [4.2.4] Algebraic Limit Theorem for Functional Limits
- [4.2.5] Divergence Criterion for Functional Limits
Solution
Let \(f: A \rightarrow \mathbb{R}\) and let \(c\) be a limit point of \(A\). We say that
$$
\begin{align*}
\lim_{x \to c} f(x) = +\infty
\end{align*}
$$
if for all \(M > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - c| < \delta\) and \(x \in A\), it follows that
$$
\begin{align*}
f(x) > M
\end{align*}
$$
So no matter how larget \(M\), we can always \(\delta\) that forces \(f(x)\) to be larger than \(M\)
To show that \(\lim\limits_{x \rightarrow 0} 1/x^2= \infty\). Then for any \(M > 0\), we want to find a \(\delta\) such that
$$
\begin{align*}
0 < |x - 0| < \delta \quad \Rightarrow \quad \frac{1}{x^2} > M
\end{align*}
$$
working backward
$$
\begin{align*}
\frac{1}{x^2} &> M \\
x^2 &< \frac{1}{M} \\
|x| &< \frac{1}{\sqrt{M}}
\end{align*}
$$
So we want to choose \(\delta\) to be \(\frac{1}{\sqrt{M}}\). So that whenever \(0 < |x - 0| < \delta\), we get
$$
\begin{align*}
|x| &< \frac{1}{\sqrt{M}} \\
x^2 &< \frac{1}{M} \\
\frac{1}{x^2} &> M
\end{align*}
$$
References
- Problem Statement Source: Aleph 0