One-Sided Limits

Section 11.2 One-Sided Limits

For one-sided limits, you will need to add an additional parameter to the limit() command, specifying which side (left or right) to approach the value from. In the case of a vertical asymptote, these limits will be equal to \(\pm \infty\text{.}\)

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> L(x) := 1/x;

\begin{equation*} \displaystyle L\, := \,x\mapsto {x}^{-1} \end{equation*}

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Aside

> plot(L(x), x=-3..3, y=-5..5);

> limit(L(x), x=0, right);

\begin{equation*} \displaystyle \infty \end{equation*}

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> limit(L(x), x=0, left);

\begin{equation*} \displaystyle -\infty \end{equation*}

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Example 11.1. Vertical Asymptotes and One-Sided Limits.

In this example, we will examine a rational function and use limits to determine its vertical asymptotes.

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> f(x) := (x^2-x-6)/(x^2-8*x+15);

\begin{equation*} \displaystyle f\, := \,x\mapsto {\frac {{x}^{2}-x-6}{{x}^{2}-8\,x+15}} \end{equation*}

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We can factor the denominator to find the domain of \(f(x)\) and predict where we might find vertical asymptotes. There is a useful denom() command for this that we can use for the denominator of \(f(x)\text{.}\)

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> factor( denom( f(x) ) );

\begin{equation*} \displaystyle \left( x-3 \right) \left( x-5 \right) \end{equation*}

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It looks like \(x=3\) and \(x=5\) are not in the domain of \(f(x)\text{,}\) though it is not clear if they are vertical asymptotes. We can find the limit of \(f(x)\) as \(x \rightarrow 3\text{.}\)

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> limit(f(x), x=3);

\begin{equation*} \displaystyle -5/2 \end{equation*}

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Since this limit exists but \(f(3)\) does not, this is a removable discontinuity and not a vertical asymptote. Now we can find the limit of \(f(x)\) as \(x \rightarrow 5\text{.}\)

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> limit(f(x), x=5);

\begin{equation*} \displaystyle \textit{undefined} \end{equation*}

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Even though this limit does not exist, we cannot automatically conclude that \(f(x)\) has a vertical asymptote at \(x=5\text{.}\) We need to compute the one-sided limits to see if there is asymptotic behaviour.

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Aside

> limit(f(x), x=5, left);

\begin{equation*} \displaystyle -\infty \end{equation*}

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> limit(f(x), x=5, right);

\begin{equation*} \displaystyle \infty \end{equation*}

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Since these limits are given as \(\pm \infty\text{,}\) we know that \(f(x)\) has a vertical asymptote at \(x=5\text{.}\)

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