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Section 1.1 The Squeeze Theorem

Motivating Questions

How can we use our knowedge of the limits of simple functions to determine the limits of more complicated functions?

Preview Activity 1.1.1

Let \(f(x)=\sin(1/x)\text{.}\)

  1. Sketch a graph of \(f(x)\) on the interval \([-1,1]\text{.}\)
  2. What is \(f(0)\text{?}\)
  3. What is \(\displaystyle\lim_{x\rightarrow 0}f(x)\,?\)

The Squeeze Theorem (also known as the Sandwich Theorem) allows us to figure out some limits that we could not otherwise, by comparing them with known limits.

Evaluating the cell below illustrates that \(-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2\) (for \(x \ne 0\)). We cannot directly compute

\begin{equation*} \lim_{x\rightarrow 0} x^2 \sin\left(\frac{1}{x}\right) \end{equation*}

because of the division by 0, but from the graph it is clear we should get 0.

The following theorem allows us to conclude that the limit really is 0.

To apply the theorem to \(\displaystyle\lim_{x\rightarrow 0} x^2 \sin\left(\frac{1}{x}\right)\text{,}\) set

\begin{align*} f(x) \amp = -x^2\,,\\ g(x) \amp = x^2 \sin\left(\frac{1}{x}\right)\,,\quad\text{and}\\ h(x) \amp= x^2\,. \end{align*}

Since \(-1 \le \sin(\cdot) \le 1\) and \(0\le x^2\text{,}\) we have \(f(x)\le g(x) \le h(x)\) when \(x\) is near \(0\) (and for all \(x\neq 0\)). We can compute

\begin{equation*} \lim_{x\rightarrow 0} f(x) =\lim_{x\rightarrow 0} h(x) = 0\text{,} \end{equation*}

so the assumptions of the Squeeze Theorem 1.1.1 are satisfied and we can conclude

\begin{equation*} \lim_{x\rightarrow 0} x^2 \sin\left(\frac{1}{x}\right) =\lim_{x\rightarrow 0} g(x) = 0\text{.} \end{equation*}
Activity 1.1.2

Let \(g(x)=x \sin\left(\frac{1}{x}\right)\) and consider the limit \(\displaystyle\lim_{x\rightarrow 0} g(x)\text{,}\) which we cannot evaluate directly.

  1. Find \(f(x)\) and \(h(x)\) such that \(f(x)\le g(x) \le h(x)\) when \(x\) is near \(0\text{.}\)
  2. Use the computational cell above or your own device to plot \(f(x)\text{,}\) \(g(x)\text{,}\) and \(h(x)\) to check that indeed \(f(x)\le g(x) \le h(x)\) when \(x\) is near \(0\text{.}\)
  3. Find \(\displaystyle\lim_{x\rightarrow a} f(x)\) and \(\displaystyle\lim_{x\rightarrow a} h(x)\text{.}\) If they are not equal, then choose new \(f(x)\) and \(h(x)\) so that they do equal.
  4. Apply the Squeeze Theorem 1.1.1 to find \(\displaystyle\lim_{x\rightarrow 0} g(x)\text{.}\)

Subsection Exercises


Use the Squeeze Theorem to find

\begin{equation*} \lim_{x\rightarrow 0} \left(2+ 3 x^2 \cos\left(\frac{1}{x}\right)\right)\,. \end{equation*}

Suppose we know \(f(x)\le g(x)\) when \(x\) is near \(a\) (except possibly at \(a\)) and \(\displaystyle\lim_{x\rightarrow a} f(x) = L\text{.}\) Can we conclude \(\displaystyle\lim_{x\rightarrow a} g(x) \ge L\text{?}\)