2.2 Limits
Let \(f:\mathbb {R}\longrightarrow \mathbb {R}\) be a given function and \(c\) and \(L\) be real numbers. Then \(f(x)\) is said to tend to a limit \(L\) as \(x\)
approaches \(c\) if for any given \(\varepsilon > 0 \hspace {0.2cm}\exists \hspace {0.2cm} \delta (\varepsilon ) > 0 \ni \) whenever \(0< \left |x - c\right |< \delta \), we have \(\left |f(x) - L\right | < \varepsilon \).
Use the definition of limits to show that \(\hspace {0.3cm}\displaystyle {\lim \limits _{x\rightarrow 3} (5x - 2) = 13}\).
Proof.
We wish to show that given any \(\varepsilon > 0\), we can find a corresponding \(\delta > 0 \ni \left |(5x -2) - 13\right | < \varepsilon \) whenever \(\hspace {0.4cm} 0 < \left |x - 3\right | < \delta \).
\begin {align*} \left |(5x - 2) - 13\right | & = \left |5x - 2 -13\right | = \left |5x - 15\right |\\ & = 5 \left |x - 3\right | \end {align*}
Therefore, given any \(\varepsilon > 0, \hspace {0.2cm} \exists \hspace {0.2cm} \delta (\varepsilon ) = \dfrac {\varepsilon }{5}> 0 \ni \) whenever \(0< \left |x - 3\right | < \delta \) we have
\begin {align*} \left |(5x - 2) - 13\right | & = 5\left |x - 3\right |\\ & < 5\delta = 5\Big (\dfrac {\varepsilon }{5}\Big )\\ & = \varepsilon \end {align*}
Therefore, \(\hspace {0.3cm}\displaystyle {\lim \limits _{x\rightarrow 3} (5x - 2) = 13}\) □
Note.
- 1.
- \(\lim \limits _{x \rightarrow c} f(x) = L \iff \) given any \(\varepsilon > 0 \hspace {0.2cm} \exists \hspace {0.2cm} \delta (\varepsilon ) > 0 \ni \left |f(x) - L\right | < \varepsilon \)
whenever \(\hspace {0.3cm} 0 < \left |x - c\right | < \delta \) - 2.
- The definition implies that there can be at most one limit.
- 3.
- The inequality \(\left |x - c\right | < \delta \implies c - \delta < x < c + \delta \). The part of the inequality which states that \(0 < \left |x - c\right |, \hspace {0.3cm} x\) is not allowed to be equal
to \(c\).
Solution.
\(\displaystyle {\left |x^2 - 4\right | = \left |(x+2)(x-2)\right | = \left |x + 2\right |\hspace {0.1cm}\left |x - 2\right |}\)
But \(x\) lies between 1 and 3 i.e \(1 < x < 3\hspace {0.3cm}\) i.e \(\left |x - 2\right | < 1\).
Now \(\left |x - 2\right | < 1 \iff 1 < x < 3\)
\(\implies \hspace {0.4cm} 2 < x + 1 < 4 \implies \left |x + 1\right | < 4\)
\(\forall \hspace {0.5cm} 1 < x < 3\)
Thus \(\hspace {0.2cm} \left |x^2 - 4\right | = \left |x + 2\right |\left |x -2\right | < 4\left |x - 2\right |\)
Therefore, given any \(\varepsilon > 0 \hspace {0.3cm} \exists \hspace {0.3cm} \delta (\varepsilon ) = \dfrac {\varepsilon }{4}> 0\)
\(\exists \hspace {0.3cm} 0 < \left |x - 2\right | < \delta \implies \left |x^2 - 4\right |< 4\left |x - 2\right | < \delta \).
i.e \(\hspace {0.3cm} \left |x^2 - 4\right | < \varepsilon \iff 0 < \left |x - 2\right | < \delta \)
Therefore, \(\hspace {0.4cm} \lim \limits _{x \rightarrow 2} x^2 = 4\).
Show directly from the definition of limits that \(\hspace {0.3cm} \lim \limits _{x \rightarrow 5} \dfrac {1}{x - 1} = \dfrac {1}{4}\)
Solution. \begin {align*} \left |f(x) - L\right | & = \left |\dfrac {1}{x - 1} - \dfrac {1}{4}\right | = \left |\dfrac {4 - (x - 1)}{4(x - 1)}\right | = \left |\dfrac {5 - x}{4(x - 1)}\right |\\\\ & = \dfrac {\left |x - 5\right | }{4\left |x - 1\right | }\\\\ & = \dfrac {1}{4}\hspace {0.1cm} \dfrac {1}{ \left |x - 1\right |}\hspace {0.1cm}\left |x - 5\right | \end {align*}
\[4 < x < 6\]
Take \(\left |x - 5\right | < 1 \iff 4 < x < 6\)
\(\implies \hspace {0.4cm} 3 < x - 1 < 5 \implies 3 < \left |x - 1\right |< 5\)
\(\implies \hspace {0.3cm} \dfrac {1}{5} < \dfrac {1}{\left |x - 1\right | } < \dfrac {1}{3}\)
\[\implies \hspace {0.5cm}\left |f(x) - L\right | = \left |\dfrac {1}{x - 1} - \dfrac {1}{4}\right | = \dfrac {1}{4}\hspace {0.1cm}\dfrac {1}{\left |x - 1\right | }\hspace {0.1cm}\left |x - 5\right |<\dfrac {1}{4}\cdot \dfrac {1}{3}= \dfrac {1}{12}\hspace {0.1cm}\left |x -5\right |\]
\(\implies \hspace {0.5cm} \left |\dfrac {1}{x - 1} - \dfrac {1}{4}\right | < \dfrac {1}{12}\hspace {0.1cm}\left |x - 5\right |\)
Therefore, given any \(\varepsilon > 0 \hspace {0.2cm} \exists \hspace {0.2cm} \delta (\varepsilon ) = 12\varepsilon \ni \) whenever \( 0 < \left |x - 5\right | < \delta \) we have \[\left |\dfrac {1}{x - 1}- \dfrac {1}{4}\right | < \dfrac {1}{12}\hspace {0.1cm}\left |x - 5\right | < \dfrac {1}{12}\delta = \frac {1}{12}(12\varepsilon ) = \varepsilon \]
Therefore, \(\hspace {0.3cm} \lim \limits _{x \rightarrow 5} \dfrac {1}{x - 1} = \dfrac {1}{4}\)
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.