Recognize when the squeeze argument fails
Bounds that stay apart
The curve is \(g(x)=|x|^p\sin\frac3x\), where the slider sets the power \(p\), and the faint dashed curves are its bounds \(|x|^p\) and \(-|x|^p\).
At \(p=0\), the bounds are the lines \(y=1\) and \(y=-1\). They are valid, which means that the curve stays between them, but they stay 2 units apart as \(x\) approaches 0. The curve crosses the whole space between them again and again.
Raise \(p\) above 0. The orange segment between the bounds now shrinks toward length 0 as \(x\) approaches 0, so the curve must approach 0 too.
Valid bounds are not enough, because the two bounds must also approach the same number.
Three conditions
The curve \(f\) lies between the bounds \(b\) and \(a\), which both approach the orange height \(\cy{L}\) at \(c\). The squeeze needs three conditions: the bounds are valid on both sides of \(c\), each has a limit, and the two limits are equal.
Press the first button. The orange segment shows a gap between the bounds at \(c\). The curve can approach any height inside the gap, so the squeeze gives no answer.
Press the second button. To the left of \(c\), the upper bound falls below the lower bound, so no curve fits between them. The red segment marks this wrong order.
Notation
The squeeze gives a limit only when all three conditions hold. First, the inequalities hold for every \(x\) near \(c\), on both sides.
\[\cq{b(x)}\le f(x)\le \cq{a(x)}\]Second, both bounds have limits at \(c\). Third, the two limits are equal.
\[\lim_{x\to c} b(x)=\lim_{x\to c} a(x)\]If a condition fails, the squeeze gives no answer. The limit of \(f\) may or may not exist, and another argument must decide.
A strict inequality uses \(\lt\) instead of \(\le\). Two functions with \(b(x)\lt a(x)\) can still have the same limit, so in the limit a strict inequality becomes non-strict, and the squeeze still works.
\[\begin{gathered}b(x)\lt f(x)\lt a(x)\\ \Downarrow\\ \lim b\le \lim f\le \lim a\end{gathered}\]To show that a limit does not exist, find two lists of inputs that approach \(c\) on which the function approaches two different numbers.
When it fails
Both curves lie between \(-1\) and 1 near 0, but the line \(y=x\) has limit 0 and \(\sin\frac3x\) has no limit.
The same bounds fit both curves, so these bounds cannot decide the limit.
Example 1: bounds that are valid but apart
Can the squeeze find \(\displaystyle\lim_{x\to 2}\cos\dfrac{5}{x-2}\)? Then decide whether the limit exists.
Because the cosine of any angle is between \(-1\) and 1, these bounds are valid for every \(x\ne2\).
\[-1\le \cos\frac{5}{x-2}\le 1\]The bounds are the constants −1 and 1, so their limits differ and the squeeze gives no answer.
\[\lim_{x\to2}(-1)=-1\ne 1=\lim_{x\to2}1\]For each positive whole number \(k\), the function equals 1 at \(x=2+\frac{5}{2\pi k}\) and \(-1\) at \(x=2+\frac{5}{(2k+1)\pi}\). Both lists of inputs approach 2 as \(k\) grows, and the orange dots mark some of them.
\[\begin{gathered}\cos(2\pi k)=1\\ \cos\big((2k+1)\pi\big)=-1\end{gathered}\]The function takes the values 1 and −1 at inputs as close to 2 as you like, so the limit does not exist.
Example 2: strict bounds
For \(x\ne 0\), the function \(f\) satisfies \(3-x^2\lt f(x)\lt 3+x^2\). Does the squeeze find \(\lim_{x\to 0}f(x)\)?
Find the limit of each bound.
\[\begin{gathered}\lim_{x\to 0}(3-x^2)=\cy{3}\\ \lim_{x\to 0}(3+x^2)=\cy{3}\end{gathered}\]A tempting wrong move: The inequalities are strict, so the limit \(L\) must satisfy a strict inequality too.
\[\cr{3\lt L\lt 3}\]No number satisfies this, yet the curve stays strictly between the bounds and still approaches 3.
In the limit, the strict inequalities become non-strict.
\[3\le L\le 3\]The only number that satisfies this is 3. The squeeze works, and \(f\) approaches 3.
\[\lim_{x\to 0}f(x)=\cy{3}\]
Practice
Can the squeeze find \(\displaystyle\lim_{x\to 0}\left(3+\sin\dfrac4x\right)\)? Decide, and say what you can conclude.
Which two horizontal lines enclose the curve? Move the point toward 0 and see whether the curve fills the space between them.
Solution
The sine is between \(-1\) and 1, so the function is between 2 and 4.
\[\cq{2}\le 3+\sin\frac4x\le \cq{4}\]The limits of the bounds are 2 and 4, so the squeeze gives no answer. At the orange dots, which approach 0, the function equals 4 or 2, so the limit does not exist.
Questions
The graph shows two functions with dashed bounds. The function \(m(x)=\cos\dfrac9x\) is near \(x=0\), and the function \(r\) is near \(x=3\).
Why can the dashed lines not find the limit of \(m(x)\) at 0?
- They approach different numbers, \(-1\) and 1
- They do not hold for \(x<0\)
- The function \(m\) has no value at 0
- The curve leaves the space between the lines
Answer
They approach different numbers, \(-1\) and 1
A student uses \(0\le r(x)\le x-3\) to find \(\displaystyle\lim_{x\to 3}r(x)\). Which statement is correct?
- The inequalities hold only for \(x>3\), so they give only the limit from the right, which is 0
- The limit is 0
- The squeeze shows that the limit does not exist
- The bounds meet at 3, so the squeeze always works
Answer
The inequalities hold only for \(x>3\), so they give only the limit from the right, which is 0
Someone claims that \(x+2\le f(x)\le 3x-2\) for all \(x\) near 2, and concludes that \(\displaystyle\lim_{x\to2}f(x)=4\). Which statement is correct?
- For x < 2, the upper bound is below the lower bound, so no function can satisfy both inequalities
- The bounds approach different numbers
- The conclusion is correct, because both bounds approach 4
- The limit does not exist
Answer
For x < 2, the upper bound is below the lower bound, so no function can satisfy both inequalities
Summary
The squeeze gives a limit only when the bounds are valid on both sides of \(c\), have limits at \(c\), and have equal limits.
If one of these conditions fails, the bounds do not decide the limit, and another argument must show whether it exists.
Check all three each time before you use the squeeze.
This is one of four skills in topic 1.8 of Which & Why: AP Calculus.