Recognize when a limit does not exist
Three ways to fail
Each button shows a function near x = 0. Move the left dot toward 0, and the right dot moves toward 0 from the other side.
The jump is \(f(x) = \lfloor x \rfloor\), the greatest integer that is at most \(x\). To the left of 0 the output is \(-1\), and to the right of 0 it is 0, so the two sides approach different numbers.
The unbounded curve is \(f(x) = \dfrac{1}{x^2}\). At \(x = 0.1\) the output is 100, and at \(x = 0.01\) it is 10,000. The outputs grow larger than any number, so the curve leaves the top of the figure.
The oscillating curve is \(f(x) = \sin\left(\dfrac1x\right)\). The outputs swing between \(-1\) and 1, faster and faster as \(x\) nears 0, so they do not approach one number.
No number works
The orange band surrounds a height \(L\) that might be the limit. Move it up or down to try other values of \(L\). Then make it narrower.
The faint vertical strip holds the inputs near \(c\). The limit is \(L\) if, however narrow the band is, all the outputs near \(c\) lie inside it. The curve is red where these outputs miss the band.
For the jump, the outputs lie at two heights, so a narrow band holds only one of them. The unbounded outputs leave every band. The oscillating outputs leave any band that is narrower than their swing.
For each curve, every L leaves some red once the band is narrow, so the limit does not exist.
Notation
A limit can fail to exist in three ways. The limits from the left and from the right can differ. The outputs can grow without bound, which means they become larger than any number. Or they can oscillate, which means they swing up and down.
When the outputs grow without bound, you may write \(\infty\) to describe the growth.
\[\lim_{x \to 0} \dfrac{1}{x^2} = \infty\]The symbol \(\infty\) is not a number, so the limit still does not exist.
In every case, the limit does not exist.
\[\lim_{x \to c} f(x) \text{ does not exist}\]A few values can mislead. Test the outputs at many inputs near \(c\), on both sides.
When it fails
If both sides rise without bound, as for \(\dfrac{1}{x^2}\), you may write \(\lim = \infty\). For \(\dfrac1x\), the left side falls without bound and the right side rises.
The two sides go in opposite directions, so neither \(\infty\) nor \(-\infty\) describes the two-sided limit. You write only the one-sided limits, and the two-sided limit does not exist.
Example 1: unbounded outputs
Find \(\lim_{x \to 2} \dfrac{3}{(x - 2)^2}\).
The denominator \((x - 2)^2\) is a square, so it stays positive as it approaches 0, and the outputs are large and positive.
At x = 2.1 and x = 1.9, the output is 300. At x = 2.01 and x = 1.99, it is 30,000. The orange bars show these outputs, which rise past the top of the figure.
\[\dfrac{3}{0.1^2} = 300, \quad \dfrac{3}{0.01^2} = 30{,}000\]The outputs grow without bound on both sides. The symbol \(\infty\) describes the growth, but the limit does not exist.
\[\lim_{x \to 2} \dfrac{3}{(x - 2)^2} = \infty\]
Example 2: oscillation
Find \(\lim_{x \to 0} \sin\left(\dfrac{\pi}{x}\right)\).
Test the inputs \(x = 1, \tfrac12, \tfrac13, \dots\) For \(x = \tfrac1n\), we have \(\pi / x = n\pi\), so the output is 0.
\[\sin(n\pi) = 0\]A tempting wrong move: All of these outputs are 0, so the limit is 0.
\[\lim_{x \to 0} \sin\left(\dfrac{\pi}{x}\right) = 0\]These inputs give only the zeros of the wave, the chalk dots. The orange dots show an output of 1 at \(x = \tfrac{2}{41}\) and \(-1\) at \(x = \tfrac{2}{43}\), and both inputs are close to 0.
Outputs of 1 and −1 occur at inputs as close to 0 as you like. No single number works, so the limit does not exist.
\[\begin{aligned}\sin\left(\dfrac{41\pi}{2}\right) &= 1, \\ \sin\left(\dfrac{43\pi}{2}\right) &= -1\end{aligned}\]
Practice
Does \(\lim_{x \to 4} \dfrac{6}{x - 4}\) exist? Describe the outputs on each side of 4.
Move the left dot toward 4. The curve is green where its outputs are positive and red where they are negative. Compare the signs on the two sides.
Solution
For \(x\) slightly greater than 4, the number \(x - 4\) is small and positive, so the outputs are large and positive. At \(x = 4.01\), the output is 600.
\[\lim_{x \to 4^+} \dfrac{6}{x - 4} = \infty\]For \(x\) slightly less than 4, the number \(x - 4\) is small and negative, so the outputs are large and negative. At \(x = 3.99\), the output is \(-600\).
\[\lim_{x \to 4^-} \dfrac{6}{x - 4} = -\infty\]The two sides go in opposite directions, so the two-sided limit does not exist. You write neither \(\infty\) nor \(-\infty\) for it.
Questions
The graph shows a function with a jump at \(x = -2\). The curve rises beside the vertical line at \(x = 1\). It is smooth at \(x = 4\).
At which input does the limit exist?
- \(x = 4\)
- \(x = -2\)
- \(x = 1\)
Answer
\(x = 4\)
At \(x = 1\) the curve rises on both sides. Which statement is correct?
- The limit does not exist. The symbol \(\infty\) only describes how the outputs grow.
- The limit exists, because both sides agree
- The limit is 0
Answer
The limit does not exist. The symbol \(\infty\) only describes how the outputs grow.
Compute \(\cos\left(\dfrac1x\right)\) at \(x = \dfrac{1}{20\pi}\) and at \(x = \dfrac{1}{21\pi}\). What do the results say about \(\lim_{x \to 0} \cos\left(\dfrac1x\right)\)?
- The limit does not exist, because the outputs are 1 and \(-1\) at inputs near 0
- The limit is 1
- The limit is 0, because the swings cancel
Answer
The limit does not exist, because the outputs are 1 and \(-1\) at inputs near 0
Summary
You test the outputs on both sides of the input and name the failure: two sides that differ, unbounded outputs, or oscillation.
This works because a limit exists only when all the outputs near the input approach one number.
Test this way when a denominator approaches 0, when a formula changes at the input, or when a function oscillates.
This is one of four skills in topic 1.2 of Which & Why: AP Calculus.