Recognize unbounded behavior on a graph
A curve beside a dashed line
The dashed vertical line marks \(x = 2\), where the curve has no point. The part of the curve above the x-axis is green, and the part below is red.
Press Approach 2 from the right. The orange point climbs: at \(x = 2.1\) the height is 30, at \(x = 2.01\) it is 300, and at \(x = 2.001\) it is 3000. The point leaves the top of the figure, so the readout gives its height.
Press Approach 2 from the left. The point falls, and at \(x = 1.9\), \(1.99\) and \(1.999\) the heights are \(-30\), \(-300\) and \(-3000\).
The heights do not approach a number. They grow without bound, which means that they pass any number you name, upward on the right and downward on the left.
A curve that does this beside a vertical line has a vertical asymptote there.
Any asymptote
A dashed vertical line stands at an input \(c\). On each side, the curve comes close to the line and leaves the figure at the top or at the bottom.
Turn each part up or down with its toggle. Each part can go either way, so there are four patterns.
Move the slider toward 0. The orange point leaves the figure on either part, because its height grows without bound.
The curve is green above the x-axis, where the outputs are positive, and red below it, where they are negative. On each side, the sign of the outputs gives the direction.
Notation
If the outputs on the right of \(c\) grow larger than any number, write \(\infty\). Green means that the outputs are positive.
\[\lim_{x \to c^+} f(x) = \cg{\infty}\]The same symbols describe the left side, with \(x \to c^-\).
If the outputs fall below any number, write \(-\infty\). Red means that the outputs are negative.
\[\lim_{x \to c^+} f(x) = \cr{-\infty}\]The symbol \(\infty\) is not a number, so a limit of \(\infty\) or \(-\infty\) describes how the limit fails to exist as a number.
The line \(x = c\) is a vertical asymptote when at least one of the one-sided limits at \(c\) is \(\infty\) or \(-\infty\).
When it fails
When the two sides go in opposite directions, no single symbol describes both. The two-sided limit does not exist and is not written as \(\infty\) or \(-\infty\).
When both sides go up, the two-sided limit is described as \(\infty\). When both go down, it is described as \(-\infty\).
Example 1: both sides climb
Find the limit from each side at \(x = -1\) for \(g(x) = \dfrac{2}{(x + 1)^2}\).
The denominator \((x + 1)^2\) is 0 at \(x = -1\), so the curve has no point there.
From the right, take \(x = -0.99\). The denominator is \(0.01^2 = 0.0001\), so \(g = 20000\), which is large and positive.
\[g(-0.99) = \cy{20000}\]From the left, take \(x = -1.01\). The denominator is \((-0.01)^2 = 0.0001\) again, because a square is never negative, so \(g = 20000\).
\[g(-1.01) = \cy{20000}\]Both sides climb, so both one-sided limits are \(\infty\), and the two-sided limit is described as \(\infty\) as well.
\[\lim_{x \to -1} g(x) = \cg{\infty}\]
Example 2: the sides go opposite ways
Find the limit from each side at \(x = -1\) for \(h(x) = \dfrac{4 - x}{x + 1}\).
Near \(x = -1\), the numerator \(4 - x\) is close to 5, which is positive, so the sign of the denominator \(x + 1\) decides the direction.
From the right, take \(x = -0.99\). The denominator is 0.01, so \(h = 4.99 / 0.01 = 499\), which is large and positive.
\[\lim_{x \to -1^+} h(x) = \cg{\infty}\]A tempting wrong move: From the left, the outputs are also large, so the limit from the left is \(\infty\) as well.
\[\lim_{x \to -1^-} h(x) = \infty\]At \(x = -1.01\) the denominator is \(-0.01\), so \(h = 5.01 / (-0.01) = -501\). The point to the left of the dashed line is on the red part, below the axis, so the curve falls there. The dashed red curve shows the climb that this move claims.
From the left, the numerator is positive and the denominator is small and negative, so the quotient is large and negative.
\[\lim_{x \to -1^-} h(x) = \cr{-\infty}\]The sides go in opposite directions, so the two-sided limit does not exist.
Practice
Find \(\lim_{x \to 1^-} p(x)\) and \(\lim_{x \to 1^+} p(x)\) for \(p(x) = -\dfrac{3}{(x - 1)^2}\).
Find the sign of the denominator on each side. Then move the slider to see which way each part goes.
Solution
The denominator \((x - 1)^2\) is small and positive on both sides, because a square is never negative.
The numerator is \(-3\), so near 1 the outputs are large and negative. At \(x = 0.99\) and \(x = 1.01\), the output is \(-3 / 0.0001 = -30000\).
\[\begin{aligned} \lim_{x \to 1^-} p(x) &= \cr{-\infty} \\ \lim_{x \to 1^+} p(x) &= \cr{-\infty} \end{aligned}\]
Questions
Use the graph for the first two questions. The dashed vertical lines mark where the curve has no point.
What is \(\lim_{x \to -2^-} f(x)\)?
- \(-\infty\)
- \(\infty\)
- \(0\)
Answer
\(-\infty\)
What is \(\lim_{x \to 3} f(x)\)?
- \(\infty\)
- \(-\infty\)
- \(3\)
Answer
\(\infty\)
What is \(\lim_{x \to 2^-} \dfrac{x - 1}{x^2 - 4}\)?
- \(-\infty\)
- \(\infty\)
- \(\tfrac{1}{4}\)
Answer
\(-\infty\)
Summary
On each side, you find the sign of the outputs and write infinity for a climb or negative infinity for a fall.
This works because the outputs grow without bound near the line, so their sign decides the direction.
Use it at a vertical asymptote on a graph, or where a denominator becomes 0 and the numerator does not.
This is one of four skills in topic 1.3 of Which & Why: AP Calculus.