Recognize an infinite discontinuity
Values that grow near x = 2
The graph shows \(f(x)=\dfrac{3}{(x-2)^2}\). The dashed vertical line marks \(x=2\), where the denominator is 0, so \(f\) has no value there.
Press Approach 2 from the left. The output grows to 300 at \(x=1.9\) and to 30,000 at \(x=1.99\). Press Approach 2 from the right. The outputs are the same, 300 at \(x=2.1\) and 30,000 at \(x=2.01\).
The outputs grow without bound on both sides, so no height is approached. The dashed line is a vertical asymptote, a vertical line near which the outputs grow without bound.
Up or down on each side
The graph now shows any function whose outputs grow without bound near an input \(c\), which the dashed line marks.
Each side of the curve goes up or down, as its orange arrow shows. Flip each side with its button.
If a side goes up, its outputs become larger than any number you name, and we write its limit as \(\infty\). If a side goes down, we write \(-\infty\).
The symbol \(\infty\) is not a number but describes how the graph behaves, so a limit written as \(\infty\) does not exist as a number.
No value at c can repair this break, because no finite height is approached.
Notation
On the left of \(c\), the curve goes up or down.
\[\lim_{x\to c^-} f(x)=\cy{\infty}\ \ \text{or}\ \ \cy{-\infty}\]Read the right side of c in the same way.
\[\lim_{x\to c^+} f(x)=\cy{\infty}\ \ \text{or}\ \ \cy{-\infty}\]If both sides go the same way, one statement describes the limit at \(c\).
\[\lim_{x\to c} f(x)=\infty\]In every case the limit does not exist as a number. The line \(x=c\) is a vertical asymptote, and the break is an infinite discontinuity.
When it fails
Send the left side down with the button. The sides now go in opposite directions, so one statement cannot describe the limit at \(c\), and you write each one-sided limit.
This happens for \(\frac{3}{x-2}\). On the left of 2 the denominator is negative, so the curve goes down. On the right it is positive, so the curve goes up. Each side is still infinite, so the break is still an infinite discontinuity.
Example 1: opposite directions
Find both one-sided limits of \(g(x)=\dfrac{5}{x+3}\) at \(x=-3\).
At \(x=-3\) the denominator is 0 but the numerator is 5, so no factor cancels and the break is not a hole. The curve is green where its outputs are positive and red where they are negative.
On the right of \(-3\), the denominator \(x+3\) is a small positive number, so \(\frac{5}{x+3}\) is large and positive. The orange point leaves the top of the figure.
\[\lim_{x\to -3^+} g(x)=\cy{\infty}\]On the left of \(-3\), the denominator is a small negative number, so \(\frac{5}{x+3}\) is large and negative. The orange point leaves the bottom of the figure.
\[\lim_{x\to -3^-} g(x)=\cy{-\infty}\]The sides go in opposite directions. The line \(x=-3\) is a vertical asymptote.
Example 2: a squared denominator
Find the limit of \(k(x)=\dfrac{x-3}{(x+1)^2}\) as \(x\) approaches \(-1\).
At \(x=-1\) the denominator is 0 but the numerator is \(-4\), so nothing cancels and \(k\) grows without bound near \(-1\).
\[-1-3=-4\]A tempting wrong move: On the left of \(-1\), the factor \(x+1\) is negative. A negative top over a negative bottom is positive, so \(k\) goes up on the left.
\[\frac{-}{-}=+\]The denominator is squared, so it is positive on both sides. At \(x=-1.01\) the value of \(k\) is \(-40{,}100\), and the red part of the curve goes down on the left.
The denominator \((x+1)^2\) is positive on both sides, and the numerator is near \(-4\), so \(k\) is large and negative on both sides.
\[\lim_{x\to -1} k(x)=\cy{-\infty}\]The limit does not exist as a number. The line \(x=-1\) is a vertical asymptote, and \(k\) has an infinite discontinuity at \(-1\).
Practice
Let \(n(x)=-\dfrac{2}{x^3}\). Find both one-sided limits at 0, and say whether the limit at 0 can be written as one statement.
Move the slider toward 0 and follow the orange points. The sign of \(x^3\) on each side explains why.
Solution
For \(x>0\) near 0, the denominator \(x^3\) is a small positive number, so \(-\frac{2}{x^3}\) is large and negative.
\[\lim_{x\to 0^+} n(x)=\cy{-\infty}\]For \(x<0\) near 0, the denominator \(x^3\) is a small negative number, so \(-\frac{2}{x^3}\) is large and positive.
\[\lim_{x\to 0^-} n(x)=\cy{\infty}\]The sides go in opposite directions, so one statement cannot describe the limit at 0. The break is an infinite discontinuity.
Questions
The graph shows a function \(F\) with dashed vertical lines at \(x=-2\) and \(x=3\).
As \(x\) approaches \(-2\), what do the values of \(F\) do?
- They go down without bound on both sides
- They go up without bound on both sides
- They go down on one side and up on the other
- They approach 0
Answer
They go down without bound on both sides
What is the right-hand limit of \(F\) as \(x\) approaches 3?
- \(\infty\)
- \(-\infty\)
- 3
- 0
Answer
\(\infty\)
Find \(\displaystyle\lim_{x\to 1^+}\frac{x-4}{x-1}\).
- \(-\infty\)
- \(\infty\)
- \(-3\)
- 0
Answer
\(-\infty\)
Summary
You decide which way the graph goes on each side of the input, and you write each side as \(\infty\) or \(-\infty\).
This works because the numerator stays near a number that is not 0 while the denominator becomes small, so the fraction becomes large.
Use it when the denominator is 0 and the numerator is not, to tell an infinite break from a hole.
This is one of five skills in topic 1.10 of Which & Why: AP Calculus.