Calculus – Limits and continuity – Limits at infinity

Definition (intuitive): Let \(f\) be a function defined on some interval \((a,\infty)\). Then

\[\lim\limits_{x\to\infty}f(x)=L\]

means that the values of \(f(x)\) can be made arbitrarily close \(L\) by requiring \(x\) to be sufficiently large.

Definition (intuitive): Let \(f\) be a function defined on some interval \((-\infty),a\). Then

\[\lim\limits_{x\to-\infty}f(x)=L\]

means that the values of \(f(x)\) can be made arbitrarily close \(L\) by requiring \(x\) to be sufficiently large negative.

Definition (precise): Let \(f\) be a function defined on some interval \((a,\infty)\). Then

\[\lim\limits_{x\to\infty}f(x)=L\]

means that for every number \(\epsilon > 0\) there is a corresponding number \(N\) such that

\[|f(x)-L| < \epsilon\quad\textrm{if}\quad x > N.\]

Definition (precise): Let \(f\) be a function defined on some interval \((-\infty,a)\). Then

\[\lim\limits_{x\to-\infty}f(x)=L\]

means that for every number \(\epsilon > 0\) there is a corresponding number \(N\) such that

\[|f(x)-L| < \epsilon\quad\textrm{if}\quad x < N.\]

In computations we often use the fact that \(\sqrt{x^2}=|x|=x\) if \(x > 0\) and \(\sqrt{x^2}=|x|=-x\) if \(x < 0\).

Examples

1) \(\displaystyle\lim\limits_{x\to\infty}\arctan(x)=\frac{1}{2}\pi\).	2) \(\displaystyle\lim\limits_{x\to-\infty}\arctan(x)=-\frac{1}{2}\pi\).
3) \(\displaystyle\lim\limits_{x\to\infty}\frac{x}{|x|}=\lim\limits_{x\to\infty}\frac{x}{x}=1\).	4) \(\displaystyle\lim\limits_{x\to-\infty}\frac{x}{|x|}=\lim\limits_{x\to-\infty}\frac{x}{-x}=-1\).
5) \(\displaystyle\lim\limits_{x\to\infty}\frac{x^2-2x+5}{x^2+3x-1}=\lim\limits_{x\to\infty}\frac{1-\displaystyle\frac{2}{x}+\frac{5}{x^2}}{1+\displaystyle\frac{3}{x}-\frac{1}{x^2}}=\frac{1-0+0}{1+0-0}=1\).	6) \(\displaystyle\lim\limits_{x\to\infty}\frac{1+2\sqrt{4x^2+1}}{2x+5}=\lim\limits_{x\to\infty}\frac{\displaystyle\frac{1}{x}+2\sqrt{4+\frac{1}{x^2}}}{2+\displaystyle\frac{5}{x}}=\frac{0+2\sqrt{4}}{2+0}=2\).
7) \(\displaystyle\lim\limits_{x\to\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}=\lim\limits_{x\to\infty}\frac{1-e^{-2x}}{1+e^{-2x}}=\frac{1-0}{1+0}=1\).	8) \(\displaystyle\lim\limits_{x\to-\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}=\lim\limits_{x\to-\infty}\frac{e^{2x}-1}{e^{2x}+1}=\frac{0-1}{0+1}=-1\).
9) \(\displaystyle\lim\limits_{x\to\infty}\left(x-\sqrt{x^2+2x}\right)=\lim\limits_{x\to\infty}\left(x-\sqrt{x^2+2x}\right)\cdot\frac{x+\sqrt{x^2+2x}}{x+\sqrt{x^2+2x}}=\lim\limits_{x\to\infty}\frac{x^2-(x^2+2x)}{x+\sqrt{x^2+2x}}=\lim\limits_{x\to\infty}\frac{-2}{1+\sqrt{1+\displaystyle\frac{2}{x}}}=-1\).
10) \(\displaystyle\lim\limits_{x\to-\infty}\left(x+\sqrt{x^2+6x}\right)=\lim\limits_{x\to-\infty}\left(x+\sqrt{x^2+6x}\right)\cdot\frac{x-\sqrt{x^2+6x}}{x-\sqrt{x^2+6x}}=\lim\limits_{x\to-\infty}\frac{x^2-(x^2+6x)}{x-\sqrt{x^2+6x}}=\lim\limits_{x\to-\infty}\frac{6}{-1-\sqrt{1+\displaystyle\frac{6}{x}}}=-3\).
11) \(\displaystyle\lim\limits_{x\to\infty}\frac{\sin(x)}{x}=0\) by the squeeze theorem, since \(-\displaystyle\frac{1}{x}\leq\frac{\sin(x)}{x}\leq\frac{1}{x}\) and \(\displaystyle\lim\limits_{x\to\infty}\left(-\frac{1}{x}\right)=0=\lim\limits_{x\to\infty}\frac{1}{x}\).
12) \(\displaystyle\lim\limits_{x\to-\infty}\arctan\left(\frac{\sqrt{x^2+4}}{x+2}\right)=\arctan\left(\lim\limits_{x\to-\infty}\frac{\sqrt{x^2+4}}{x+2}\right)=\arctan\left(\lim\limits_{x\to-\infty}\frac{\sqrt{1+\displaystyle\frac{4}{x^2}}}{-1-\displaystyle\frac{2}{x}}\right)=\arctan(-1)=-\frac{1}{4}\pi\).

---

Last modified on October 26, 2021