Check one-sided or domain restrictions
A formula that changes at 5
The graph shows \(g(x)=\dfrac{x^2-25}{|x-5|}\). The bars mean absolute value, which is the size of \(x-5\) without its sign. The orange dot sits on the curve at the input x.
Move the dot toward 5 from the right. Here x is greater than 5, so \(|x-5|=x-5\), the quotient simplifies to \(x+5\), and the dot approaches height 10.
Move the dot toward 5 from the left. Here x is less than 5, so \(|x-5|=5-x\), the quotient simplifies to \(-(x+5)\), and the dot approaches height \(-10\).
The two sides approach different heights, because the direction of approach changes the formula.
Two sides, one limit
The curve is now any function with a break at \(c\). Two orange dots approach \(c\), one from each side. The left dot uses inputs less than \(c\), and the right dot uses inputs greater than \(c\).
The heights that the dots approach are the left-hand limit \(\cy{L}\) and the right-hand limit \(\cy{R}\). When they differ, the red bar shows the gap between them.
Press Make L equal R. Both dots then approach the same height, so the two-sided limit exists.
Press Remove the left side. The left dot disappears because the function has no values there, so only the right-hand limit remains.
Notation
The notation \(x\to c^-\) means that \(x\) approaches \(c\) from the left, through values less than \(c\). The notation \(x\to c^+\) means that \(x\) approaches \(c\) from the right.
\[\cy{\lim_{x\to c^-}f(x)}\qquad \cy{\lim_{x\to c^+}f(x)}\]The two-sided limit exists when both one-sided limits exist and are equal. If they differ, as the red bar shows here, the two-sided limit does not exist.
\[\begin{aligned}\lim_{x\to c^-}f(x)&=L\\ \lim_{x\to c^+}f(x)&=L\\ \Longrightarrow\ \lim_{x\to c}f(x)&=L\end{aligned}\]An absolute value has a different formula on each side of the input where its inside is 0. Write it without the bars on each side.
\[|u|=\begin{cases}u & u\ge 0\\ -u & u<0\end{cases}\]A limit uses only the inputs where the function is defined. At the end of a domain, only one side has inputs, so the two-sided limit does not exist there.
On each side, use the usual methods: substitute, factor or multiply by the conjugate.
When it fails
If the left side is hidden, press Restore the left side. Move the slider away from 0, and the red bar shows the gap between the two heights.
No single number is approached from both sides, so the limit does not exist. One side alone cannot decide it.
Example 1: the end of a domain
Let \(f(x)=\sqrt{x-6}\). Find the left-hand and right-hand limits of \(f\) as \(x\) approaches 6, and decide whether the two-sided limit exists.
The square root is defined only when \(x-6\ge 0\), that is, when \(x\ge 6\). The graph has no points to the left of 6, and the faint region marks those inputs.
\[x-6\ge 0 \iff x\ge 6\]From the right, \(x-6\) approaches 0 through positive values, so the square root approaches 0.
\[\lim_{x\to 6^+}\sqrt{x-6}=\cy{0}\]From the left there are no inputs, so the left-hand limit does not exist.
\[\lim_{x\to 6^-}\sqrt{x-6}\ \text{does not exist}\]The two-sided limit needs both sides, so it does not exist at 6.
Example 2: a formula with 0 over 0 on one side
Let \(k(x)=\dfrac{x^2-9}{x-3}\) for \(x<3\) and \(k(x)=2x\) for \(x\ge 3\). Find \(\displaystyle\lim_{x\to 3}k(x)\).
Find each side with its own formula. On the left, substituting \(x=3\) gives 0 on the top and 0 on the bottom.
\[\frac{3^2-9}{3-3}=\frac00\]A tempting wrong move: The left side gives \(\dfrac00\), so the left-hand limit does not exist.
\[\lim_{x\to 3^-}k(x)\ \text{does not exist}\]The left part ends at an open circle at height 6. The form \(\dfrac00\) does not mean that the limit is missing. It means that a rewrite is needed.
Factor the top and cancel \(x-3\), because \(x\) is not 3 in a limit. For \(x<3\), the formula equals \(x+3\).
\[\begin{gathered}\lim_{x\to 3^-}\frac{(x-3)(x+3)}{x-3}\\ =\lim_{x\to 3^-}(x+3)=\cy{6}\end{gathered}\]On the right, \(k(x)=2x\) approaches 6. Both one-sided limits are 6, so the limit is 6.
\[\lim_{x\to 3^+}2x=\cy{6}\]
Practice
Find the left-hand and right-hand limits of \(\dfrac{|x+4|}{x^2-16}\) as \(x\) approaches \(-4\). Decide whether the limit exists.
Write \(|x+4|\) without the bars on each side of \(-4\). The graph has a separate piece on each side.
Solution
Factor the bottom as \((x-4)(x+4)\). On the right, \(x+4>0\), so \(|x+4|=x+4\).
\[\begin{gathered}\lim_{x\to -4^+}\frac{x+4}{(x-4)(x+4)}\\ =\lim_{x\to -4^+}\frac{1}{x-4}=\cy{-\frac18}\end{gathered}\]On the left, \(x+4<0\), so \(|x+4|=-(x+4)\).
\[\begin{gathered}\lim_{x\to -4^-}\frac{-(x+4)}{(x-4)(x+4)}\\ =\lim_{x\to -4^-}\frac{-1}{x-4}=\cy{\frac18}\end{gathered}\]The one-sided limits are different, so the limit as \(x\) approaches \(-4\) does not exist.
Questions
The graph shows two curves. The left curve is \(u(x)\), with an open circle at \((1,3)\) and a filled dot at \((1,1)\). The right curve is \(v(x)=\sqrt{x-4}\), which starts at \((4,0)\).
What is the left-hand limit of \(u(x)\) as \(x\) approaches 1?
- 3
- 1
- 2
- It does not exist
Answer
3
Which statement about \(v(x)=\sqrt{x-4}\) at \(x=4\) is correct?
- The right-hand limit is 0, and the two-sided limit does not exist
- The two-sided limit is 0, because v(4) = 0
- The left-hand limit is 0, because v(4) = 0
- No limit of any kind exists, because the curve starts at 4
Answer
The right-hand limit is 0, and the two-sided limit does not exist
Let \(g(x)=x^2-3\) for \(x<2\), \(g(x)=3x-5\) for \(x>2\), and \(g(2)=7\). Find \(\displaystyle\lim_{x\to 2}g(x)\).
- 1
- 7
- It does not exist, because the formula changes at 2
Answer
1
Summary
You find the limit from the left and from the right, each with the formula that applies on that side.
The two-sided limit exists when both sides approach the same number.
Find each one-sided limit when the formula changes at the input or the domain ends there.
This is one of four skills in topic 1.7 of Which & Why: AP Calculus.