Check that the bounds meet
Two bounds that close in on a point
A function \(f\) is known only through two inequalities, which hold for \(x\) near 2: \(4x-3\le f(x)\le x^2+1\). The dashed line is the lower bound, the dashed parabola is the upper bound, and the chalk curve \(f\) stays between them.
Move x toward 2. The orange segment joins the two bounds at the input x, so its length is the gap between them.
At x = 2.2, the bounds are 5.8 and 5.84, so the gap is 0.04. As x approaches 2, the gap shrinks to 0 and both bounds approach 5.
The curve \(f\) lies in the gap, so \(f\) must approach 5 too. This argument is called the squeeze theorem.
When the gap closes at c
The curves are now any bounds \(b\) and \(a\), with \(f\) between them. The two orange dots on the line at \(c\) mark the height that each bound approaches.
With the slider at 0, the dots coincide, so the gap closes at c and f is forced to that height.
Move the slider up. The dots separate, so a gap stays open at c, and f can approach any height inside it.
The squeeze theorem works only when both bounds approach the same number.
Notation
Find the limit of the lower bound as \(x\) approaches \(c\).
\[\cy{\lim_{x\to c} b(x)}\]Find the limit of the upper bound as \(x\) approaches \(c\).
\[\cy{\lim_{x\to c} a(x)}\]If the two limits are the same number \(L\), then \(f\) approaches \(L\), because the gap that holds \(f\) shrinks to one height.
\[\begin{gathered}\lim_{x\to c} b(x)=L=\lim_{x\to c} a(x)\\ \Downarrow\\ \lim_{x\to c} f(x)=\cy{L}\end{gathered}\]If the two limits are different numbers, the squeeze gives no answer.
The squeeze also needs \(b(x)\le f(x)\le a(x)\) for \(x\) near \(c\). Each bound is an ordinary function, so you find its limit with the usual methods, and both limits must exist.
When it fails
Move the slider up. A gap stays open at \(c\), and two green curves \(f_1\) and \(f_2\), each a possible \(f\), fit between the bounds.
The curve \(f_1\) approaches a lower height in the gap than \(f_2\). Both obey the inequalities, so the bounds do not decide the limit.
Example 1: bounds that meet at 6
For \(x\ne 0\), the function \(g\) satisfies \(6-x^2\le g(x)\le 6+2x^2\). Find \(\displaystyle\lim_{x\to 0}g(x)\).
The lower bound approaches 6, because \(x^2\) approaches 0.
\[\lim_{x\to 0}(6-x^2)=\cy{6}\]The upper bound also approaches 6, because \(2x^2\) approaches 0.
\[\lim_{x\to 0}(6+2x^2)=\cy{6}\]Both limits are 6, so the bounds meet. The function \(g\) is between them, so \(g\) approaches 6.
\[\lim_{x\to 0}g(x)=\cy{6}\]
Example 2: only one bound checked
For \(x\) near 1, the function \(f\) satisfies \(2x+1\le f(x)\le x^2+4\). What does the squeeze say about \(\lim_{x\to 1}f(x)\)?
Find the limit of the upper bound.
\[\lim_{x\to 1}(x^2+4)=\cy{5}\]A tempting wrong move: The upper bound approaches 5, so \(f\) approaches 5.
\[\lim_{x\to 1}f(x)=\cr{5}\]The lower bound is at height 3 above \(x=1\), so a gap from 3 to 5 stays open. The green curves both fit in it but pass through 3.5 and 4.5 there, so \(f\) need not approach the red 5.
Find the limit of the lower bound too.
\[\lim_{x\to 1}(2x+1)=\cy{3}\]The limits are 3 and 5, so the bounds do not meet. The squeeze gives no answer, because f can approach any height between them.
Practice
For \(x\) near 3, the function \(h\) satisfies \(6x-12\le h(x)\le x^2-3\). Find \(\displaystyle\lim_{x\to 3}h(x)\).
Read the height of each bound above x = 3 on the figure. Do the two curves meet there?
Solution
Both bounds are polynomials, so substitute x = 3.
\[\lim_{x\to 3}(6x-12)=6\cdot 3-12=\cy{6}\]The upper bound approaches 6 too.
\[\lim_{x\to 3}(x^2-3)=9-3=\cy{6}\]The bounds meet at 6, so \(h\) approaches 6.
\[\lim_{x\to 3}h(x)=\cy{6}\]
Questions
The graph shows two pairs of bounds. One pair is near \(x=1\), and the other pair is near \(x=-2\).
As \(x\) approaches 1, which height do both bounds approach?
- 2
- 1
- 3
- They approach different heights
Answer
2
A function lies between the two bounds near \(x=-2\). What does the squeeze say about its limit as \(x\) approaches \(-2\)?
- Nothing, because the bounds approach 0 and 3
- The limit is 0
- The limit is 3
- The limit is 1.5
Answer
Nothing, because the bounds approach 0 and 3
For \(x\) near 0, \(5-3x^2\le f(x)\le 5\cos x\). Find \(\displaystyle\lim_{x\to 0}f(x)\).
- 5
- 2
- 0
- The limit cannot be found
Answer
5
Summary
You find the limit of each bound and compare the two limits.
If they are the same number, the gap that holds f shrinks to one height, so f approaches that height.
Check this before you state any squeeze limit.
This is one of four skills in topic 1.8 of Which & Why: AP Calculus.