Lesson 1.6 · Unit 1 · CED topic 1.8
Determining limits using the squeeze theorem
Some functions refuse every algebraic trick. Try factoring \(x^2\sin\dfrac1x\) or multiplying it by a conjugate and you get nowhere, and direct substitution at \(x=0\) is not even possible. But such a function has a feature we can exploit: it is trapped between two simpler functions that are heading to the same place. If a function is pinned between two others that both approach \(L\), it has no choice but to approach \(L\) as well. That is the squeeze theorem.
Learning objective.
- LIM-1.E. Determine the limits of functions using equivalent expressions for the function or the squeeze theorem.
Essential knowledge.
- LIM-1.E.2. The limit of a function may be found by using the squeeze theorem.
The theorem
Squeeze theorem. Suppose that \(g(x)\le f(x)\le h(x)\) for all \(x\) near \(c\) (except possibly at \(x=c\) itself), and that \[\lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L.\] Then \(\displaystyle\lim_{x\to c}f(x)=L\).
This is a theorem, so on the exam you must check its hypotheses before you use it: the inequality really holds near \(c\), and the two bounding limits exist and are equal. Then name the theorem and state the conclusion.
In the figure, \(f(x)=x^2\sin\dfrac1x\) oscillates wildly near \(0\), but because \(-1\le\sin\dfrac1x\le1\) we have \(-x^2\le f(x)\le x^2\). The outer curves both pass through the origin, and the oscillating curve is forced through it too.
Building the bounds
Almost every squeeze problem starts with a bounded function such as \(\sin\), \(\cos\), or \(\sin\dfrac1x\), whose values always lie between \(-1\) and \(1\). Start from the bound on the bounded part, then multiply through by the other factor. If that factor can be negative, multiply by its absolute value or square (which is nonnegative) so the inequality direction is preserved.
Find \(\displaystyle\lim_{x\to0}x^4\cos\frac{2}{x}\).
Bound. For all \(x\ne0\), \(-1\le\cos\dfrac2x\le1\). Since \(x^4\ge0\), multiplying keeps the direction: \(-x^4\le x^4\cos\dfrac2x\le x^4\).
Limits of the bounds. \(\displaystyle\lim_{x\to0}(-x^4)=0\) and \(\displaystyle\lim_{x\to0}x^4=0\).
Conclusion. By the squeeze theorem, \(\displaystyle\lim_{x\to0}x^4\cos\frac2x=0\).
For all \(x\) near 2 (with \(x\ne2\)), a function \(f\) satisfies \(4x-5\le f(x)\le x^2-1\). Find \(\displaystyle\lim_{x\to2}f(x)\).
\(\displaystyle\lim_{x\to2}(4x-5)=3\) and \(\displaystyle\lim_{x\to2}(x^2-1)=3\). The bounds agree, so by the squeeze theorem \(\displaystyle\lim_{x\to2}f(x)=3\). We learned this without knowing a formula for \(f\).
The limit \(\dfrac{\sin x}{x}\)
The squeeze theorem is the reason \(\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1\), the limit you used in Lesson 1.6. Compare areas in the unit circle for \(0\lt x\lt\dfrac\pi2\): the triangle with height \(\sin x\) is inside the sector of angle \(x\), which is inside the triangle with height \(\tan x\). This gives \(\sin x\lt x\lt\tan x\). Dividing by \(\sin x\) and taking reciprocals:
\[\cos x\lt\frac{\sin x}{x}\lt1.\]
Because \(\dfrac{\sin x}{x}\) is an even function, the same inequality holds for negative \(x\) near \(0\). As \(x\to0\), \(\cos x\to1\) and the upper bound is \(1\), so \(\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1\) by the squeeze theorem.
A student argues: "\(\sin\dfrac1x\) has no limit as \(x\to0\), so \(x\sin\dfrac1x\) has no limit either." Is that right?
No. The product of a function with no limit and a function that approaches \(0\) can have a limit. Here \(-|x|\le x\sin\dfrac1x\le|x|\), and both bounds approach \(0\), so \(\displaystyle\lim_{x\to0}x\sin\dfrac1x=0\) by the squeeze theorem. Compare \(\sin\dfrac1x\) alone, which does not settle at all.
- State the hypotheses: \(g(x)\le f(x)\le h(x)\) near \(c\), and \(\lim g=\lim h=L\).
- Bound a trig factor with \(-1\le\sin\le1\) or \(-1\le\cos\le1\), then multiply by a nonnegative factor (use \(x^2\), \(x^4\), or \(|x|\)).
- If the two bounding limits are different, the squeeze theorem says nothing.
- \(\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1\) is itself a squeeze result.
- Use the squeeze theorem to evaluate \(\displaystyle\lim_{x\to0}x^2\cos\dfrac{5}{x}\).
Show answer
Since \(-1\le\cos\dfrac5x\le1\) and \(x^2\ge0\), \(-x^2\le x^2\cos\dfrac5x\le x^2\). Both bounds approach 0, so the limit is 0.
- For all \(x\) near 1, \(2x\le f(x)\le x^2+1\). Find \(\displaystyle\lim_{x\to1}f(x)\), and check that the bounds are valid.
Show answer
\(\displaystyle\lim_{x\to1}2x=2\) and \(\displaystyle\lim_{x\to1}(x^2+1)=2\), so \(\displaystyle\lim_{x\to1}f(x)=2\) by the squeeze theorem. The inequality \(2x\le x^2+1\) holds because \(x^2-2x+1=(x-1)^2\ge0\).
- Evaluate \(\displaystyle\lim_{x\to0}x^2\left(2+\sin\dfrac1x\right)\).
Show answer
\(1\le2+\sin\dfrac1x\le3\), so \(x^2\le x^2\left(2+\sin\dfrac1x\right)\le3x^2\). Both bounds approach 0, so the limit is 0.
- A student says: "\(g(x)\le f(x)\le h(x)\), \(\lim g=1\) and \(\lim h=3\), so by the squeeze theorem \(\lim f=2\)." What is wrong?
Show answer
The squeeze theorem needs the two bounding limits to be equal. Here \(1\ne3\), so the theorem gives no information; \(f\) could approach any number between 1 and 3 (or fail to have a limit at all).
- Show that \(\displaystyle\lim_{x\to0}x\sin\dfrac1x=0\) even though \(\displaystyle\lim_{x\to0}\sin\dfrac1x\) does not exist.
Show answer
\(-1\le\sin\dfrac1x\le1\) gives \(-|x|\le x\sin\dfrac1x\le|x|\); both \(\pm|x|\) approach 0, so the limit is 0. The function \(\sin\dfrac1x\) alone oscillates between \(-1\) and \(1\) near 0 and has no limit.
- Using \(\cos x\lt\dfrac{\sin x}{x}\lt1\) for small positive \(x\), explain why \(\displaystyle\lim_{x\to0^+}\dfrac{\sin x}{x}=1\).
Show answer
As \(x\to0^+\), \(\cos x\to1\) and the upper bound is the constant 1. Both bounds approach 1, so \(\dfrac{\sin x}{x}\) is squeezed to 1.