Use the bounds to find the limit
A wobble that shrinks
The curve is \(f(x)=4+(x-2)\cos\frac{6}{x-2}\). It wobbles around the height 4, faster and faster near \(x=2\), and the cosine part has no limit there.
The cosine is between \(-1\) and 1, so the wobble \((x-2)\cos\frac{6}{x-2}\) is at most \(|x-2|\) in size. This means that \(f(x)\) lies between \(4-|x-2|\) and \(4+|x-2|\), which are the faint dashed V shapes.
Drag the orange point toward 2. The orange segment between the bounds shrinks toward the height 4, because |x − 2| approaches 0. At x = 2.2, it runs from 3.8 to 4.2. The second button draws the height 4 as a dashed orange line.
The curve stays between the bounds, so \(f(x)\) approaches 4.
A pinch toward a steady value
The curve is a steady value \(\cy{L}\) plus a wobble, which is a size \(s\) times a wave \(w\) that stays between \(-1\) and 1.
The faint curves \(L-s\) and \(L+s\) bound the curve. Drag the orange point toward \(c\). The size \(s\) shrinks, so the bounds close in on the orange line at \(L\). The slider sets how fast \(s\) shrinks.
The wave can have any shape inside the bounds. When the size \(s\) approaches 0, the curve approaches \(L\).
The value \(L\) does not have to be 0. It is the common limit of the two bounds.
Notation
Suppose \(f\) is between a lower bound \(b\) and an upper bound \(a\) near \(c\).
\[\cq{b(x)}\le f(x)\le \cq{a(x)}\]By the squeeze theorem, if both bounds approach the same number \(L\), then \(f\) approaches \(L\).
\[\begin{gathered}\lim_{x\to c} b(x)=L=\lim_{x\to c} a(x)\\ \Downarrow\\ \lim_{x\to c} f(x)=\cy{L}\end{gathered}\]A common pattern is a steady value \(L\) plus a size \(s(x)\) times a wave \(w(x)\), where \(|w(x)|\le 1\). Because \(|s\,w|\) is at most \(|s|\), the bounds are \(L-|s|\) and \(L+|s|\).
\[\cq{L-|s|}\le L+s\,w\le \cq{L+|s|}\]If \(|s(x)|\) approaches 0, both bounds approach \(L\), so the whole expression approaches \(L\).
The wave \(w\) may have no limit, but this does not matter, because the argument never takes the limit of \(w\).
When it fails
Press the toggle to keep the size \(s\) the same. The bounds stay apart, and the curve continues to oscillate between them.
The squeeze theorem needs the size to approach 0, because otherwise the bounds do not meet and give no answer.
Example 1: a factor that stays positive
Find \(\displaystyle\lim_{x\to 0}x^2\left(2+\sin\dfrac5x\right)\).
The sine is between \(-1\) and 1, so \(2+\sin\frac5x\) is between 1 and 3.
\[1\le 2+\sin\frac5x\le 3\]Multiply by \(x^2\), which is never negative, so the inequalities keep their direction. These bounds are the faint dashed parabolas.
\[\cq{x^2}\le x^2\left(2+\sin\tfrac5x\right)\le \cq{3x^2}\]Both bounds approach 0 as \(x\) approaches 0, so the function approaches 0.
\[\lim_{x\to 0}x^2\left(2+\sin\tfrac5x\right)=\cy{0}\]
Example 2: a factor with no limit
Find \(\displaystyle\lim_{x\to 3}(x-3)\sin\dfrac{2}{x-3}\).
The factor \(\sin\frac{2}{x-3}\) oscillates between \(-1\) and 1 infinitely often near 3, so it has no limit at 3.
A tempting wrong move: One factor has no limit, so the product has no limit.
\[\cr{\text{no limit}}\]The orange band shows the heights within 0.1 of 0, and the faint strip shows the inputs within 0.1 of 3. Inside the strip, the curve never leaves the band. The oscillations are real, but they shrink toward 0.
Bound the product instead. The sine is at most 1 in size, so the product is at most \(|x-3|\) in size and lies between the faint dashed V shapes.
\[\begin{gathered}|f(x)|=|x-3|\left|\sin\tfrac{2}{x-3}\right|\\ |f(x)|\le |x-3|\\ -|x-3|\le f(x)\le |x-3|\end{gathered}\]Both bounds approach 0 as \(x\) approaches 3, from either side, so \(f\) approaches 0.
\[\lim_{x\to 3}(x-3)\sin\tfrac{2}{x-3}=\cy{0}\]
Practice
Find \(\displaystyle\lim_{x\to 1}\left(x^2+(x-1)\cos\dfrac{4}{x-1}\right)\).
Drag the point toward 1, where the swings shrink. Bound the cosine term by \(|x-1|\), as in Example 2.
Solution
The cosine is between \(-1\) and 1, so the term \((x-1)\cos\frac{4}{x-1}\) is at most \(|x-1|\) in size. Adding \(x^2\) gives the faint dashed bounds.
\[\cq{x^2-|x-1|}\le f\le \cq{x^2+|x-1|}\]As \(x\) approaches 1, \(x^2\) approaches 1 and \(|x-1|\) approaches 0, so both bounds approach 1.
\[\lim_{x\to 1}\left(x^2\pm|x-1|\right)=\cy{1}\]The function is between the bounds, so it approaches 1.
\[\lim_{x\to 1}f(x)=\cy{1}\]
Questions
The graph shows two functions with their dashed bounds. The function \(q(x)=|x|\left(4+3\cos\dfrac1x\right)\) is near \(x=0\), and \(p(x)=3+(x-3)\cos\dfrac{8}{x-3}\) is near \(x=3\).
As \(x\) approaches 3, what does \(p(x)\) approach?
- 3
- 4
- 0
- It does not exist, because the cosine oscillates
Answer
3
As \(x\) approaches 0, what does \(q(x)\) approach?
- 0
- 7
- 4
- It does not exist, because the cosine oscillates
Answer
0
Find \(\displaystyle\lim_{x\to 0}\left(4+x^3\sin\dfrac{1}{x^2}\right)\).
- 4
- 5
- 0
- The limit does not exist
Answer
4
Summary
To find the limit, you write the function between two bounds and find the limit of each bound.
If both bounds approach the same number L, the function approaches L too, even when it oscillates.
The method fits a factor that approaches 0 times a bounded oscillating factor, such as a sine.
This is one of four skills in topic 1.8 of Which & Why: AP Calculus.