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11.22 Sequential Limit

Limits at infinity and infinite limits

Example

\(f(x) = \frac{1}{x^2}\)

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Limits at infinity: \(\lim_{x \to \infty} \frac{1}{x^2} = 0\) \(\lim_{x \to -\infty} \frac{1}{x^2} = 0\)

Infinite limits: \(\lim_{x \to 0^+} \frac{1}{x^2} = \infty\) \(\lim_{x \to 0^-} \frac{1}{x^2} = \infty\)

Definitions

\(\lim_{x \to +\infty} f(x) = L\) if \(\forall \epsilon > 0\), \(\exists N > 0\) such that \(\forall x > N\), it follows that \(|f(x) - L| < \epsilon\).

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\(\lim_{x \to -\infty} f(x) = L\) if \(\forall \epsilon > 0\), \(\exists N < 0\) such that \(\forall x < N\), we have that \(|f(x) - L| < \epsilon\).

Example

Prove that \(\lim_{x \to \infty} \frac{1}{x} = 0\)

To prove \(\lim_{x \to \infty} \frac{1}{x} = 0\), I need to prove that for \(\epsilon > 0\), \(\exists N > 0\) such that \(\forall x > N\), we have \(|\frac{1}{x} - 0| < \epsilon\).

For \(\epsilon > 0\), we must find \(N\) with all the requirements.

\(|f(x) - L| = \left| \frac{1}{x} - 0 \right| = \left| \frac{1}{x} \right| < \epsilon\)

Then \(0 < \frac{1}{x} < \epsilon \implies x > \frac{1}{\epsilon}\)

Taking \(N = \frac{1}{\epsilon}\), the definition holds.

Examples

  1. \(\lim_{x \to \infty} \frac{x^2 - x + 1}{2x^2 - x + \pi} = \lim_{x \to \infty} \frac{x^2 \left(1 - \frac{1}{x} + \frac{1}{x^2}\right)}{x^2 \left(2 - \frac{1}{x} + \frac{\pi}{x^2}\right)} = \lim_{x \to \infty} \frac{1 - \frac{1}{x} + \frac{1}{x^2}}{2 - \frac{1}{x} + \frac{\pi}{x^2}} = \lim_{x \to \infty} \frac{1}{2} = \frac{1}{2}\)

    \(\lim_{x \to \infty} \frac{P(x)}{Q(x)} = \frac{a_n}{b_n}\) when \(\deg(P) = \deg(Q)\). Here \(P(x) = a_n x^n + \dots + a_0, Q(x) = b_n x^n + \dots + b_0\)​

    \(\lim_{x\to\infty}\frac{P(x)}{Q(x)}=\begin{cases}\frac{a_{n}}{b_{n}}, & \text{if }\deg(P)=\deg(Q),\\ 0, & \text{if }\deg(P)<\deg(Q),\\ \infty, & \text{if }\deg(P)>\deg(Q)\text{ and }\frac{a_{n}}{b_{m}}>0,\\ -\infty, & \text{if }\deg(P)>\deg(Q)\text{ and }\frac{a_{n}}{b_{m}}<0.\end{cases}\)​

  2. \(\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right) = \infty - \infty\) (indeterminate form)

    \(=\lim_{x\to\infty}\left(\sqrt{x^2+x}-x\right)\cdot\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty}\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\lim_{x\to\infty}\frac{x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac12\)​

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