Construct a definition for the statement of the form \(\lim\limits_{x \rightarrow \infty} f(x) = L\) and use it to prove the \(\lim\limits_{x \rightarrow \infty} 1/x = 0\)
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 \(f: A \rightarrow \mathbb{R}\). We say that
$$
\begin{align*}
\lim_{x \to \infty} f(x) = L
\end{align*}
$$
if for all \(\epsilon > 0\), there exists an \(N > 0\) such that whenever \(x > N\) and \(x \in A\), it follows that
$$
\begin{align*}
|f(x) - L | < \epsilon
\end{align*}
$$
To show that \(\lim\limits_{x \rightarrow \infty} 1/x = 0\). Then for any \(\epsilon > 0\), we want to find an \(\N\) such that
$$
\begin{align*}
x > N \quad \Rightarrow \quad |\frac{1}{x} - 0| < \epsilon
\end{align*}
$$
working backward
$$
\begin{align*}
\frac{1}{x} &< \epsilon \\
x &> \frac{1}{\epsilon}
\end{align*}
$$
So we want to choose \(N\) to be \(\frac{1}{\epsilon}\). So that whenever \(x > N\), we get
$$
\begin{align*}
x &> \frac{1}{\epsilon} \\
\frac{1}{x} &< \epsilon \\
|\frac{1}{x} - 0| &< \epsilon
\end{align*}
$$
References
- Problem Statement Source: Aleph 0