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Section 3.6 The Squeeze Theorem

In this section we aim to compute the limit:

\begin{equation*} \lim_{x\to0} {\sin x\over x}\text{.} \end{equation*}

We start by analyzing the graph of \(\ds{y=\frac{\sin x}{x}}\text{:}\)

Notice that \(x=0\) is not in the domain of this function. Nevertheless, we can look at the limit as \(x\) approaches \(0\text{.}\) From the graph we find that the limit is \(1\) (there is an open circle at \(x=0\) indicating \(0\) is not in the domain).

We just convinced you this limit formula holds true based on the graph, but how does one attempt to prove this limit more formally? To do this we need to be quite clever, and to employ some indirect reasoning. The indirect reasoning is embodied in a theorem, frequently called the Squeeze Theorem.

This theorem can be proved using the official definition of limit. We won't prove it here, but point out that it is easy to understand and believe graphically. The condition says that \(f(x)\) is trapped between \(g(x)\) below and \(h(x)\) above, and that at \(x=a\text{,}\) both \(g\) and \(h\) approach the same value. This means the situation looks something like Figure 3.2.

For example, imagine the blue curve is \(f(x)=x^2\sin(\pi/x)\text{,}\) the upper (red) and lower (green) curves are \(h(x)=x^2\) and \(g(x)=-x^2\text{.}\) Since the sine function is always between \(-1\) and \(1\text{,}\) \(-x^2\le x^2\sin(\pi/x)\le x^2\text{,}\) and it is easy to see that \(\lim_{x\to0}-x^2=0=\lim_{x\to0}x^2\text{.}\) It is not so easy to see directly (i.e. algebraically) that \(\lim_{x\to0}x^2\sin(\pi/x)=0\text{,}\) because the \(\pi/x\) prevents us from simply plugging in \(x=0\text{.}\) The Squeeze Theorem makes this “hard limit” as easy as the trivial limits involving \(x^2\text{.}\)

Figure 3.2. The Squeeze Theorem.

To compute \(\ds\lim_{x\to0} (\sin x)/x\text{,}\) we will find two simpler functions \(g\) and \(h\) so that \(g(x)\le (\sin x)/x\le h(x)\text{,}\) and so that \(\lim_{x\to0}g(x)=\lim_{x\to0}h(x)\text{.}\) Not too surprisingly, this will require some trigonometry and geometry. Referring to Figure, \(x\) is the measure of the angle in radians. Since the circle has radius 1, the coordinates of point \(A\) are \((\cos x,\sin x)\text{,}\) and the area of the small triangle is \((\cos x\sin x)/2\text{.}\) This triangle is completely contained within the circular wedge-shaped region bordered by two lines and the circle from \((1,0)\) to point \(A\text{.}\) Comparing the areas of the triangle and the wedge we see \((\cos x\sin x)/2 \le x/2\text{,}\) since the area of a circular region with angle \(\theta\) and radius \(r\) is \(\theta r^2/2\text{.}\) With a little algebra this turns into \((\sin x)/x \le 1/\cos x\text{,}\) giving us the \(h\) we seek.

To find \(g\text{,}\) we note that the circular wedge is completely contained inside the larger triangle. The height of the triangle, from \((1,0)\) to point \(B\text{,}\) is \(\tan x\text{,}\) so comparing areas we get \(x/2 \le (\tan x)/2 = \sin x / (2\cos x)\text{.}\) With a little algebra this becomes \(\cos x \le (\sin x)/x\text{.}\) So now we have

\begin{equation*} \cos x \le {\sin x\over x}\le {1\over\cos x}\text{.} \end{equation*}

Finally, the two limits \(\lim_{x\to0}\cos x\) and \(\lim_{x\to0}1/\cos x\) are easy, because \(\cos(0)=1\text{.}\) By the Squeeze Theorem, \(\lim_{x\to0} (\sin x)/x = 1\) as well.

Using the above, we can compute a similar limit:

\begin{equation*} \lim_{x\to0}{\cos x - 1\over x}\text{.} \end{equation*}

This limit is just as hard as \(\sin x/x\text{,}\) but closely related to it, so that we don't have to do a similar calculation; instead we can do a bit of tricky algebra.

\begin{equation*} {\cos x - 1\over x}={\cos x - 1\over x}{\cos x+1\over\cos x+1} ={\cos^2 x - 1\over x(\cos x+1)}={-\sin^2 x\over x(\cos x+1)}= -{\sin x\over x}{\sin x\over \cos x + 1}\text{.} \end{equation*}

To compute the desired limit it is sufficient to compute the limits of the two final fractions, as \(x\) goes to 0. The first of these is the hard limit we've just done, namely 1. The second turns out to be simple, because the denominator presents no problem:

\begin{equation*} \lim_{x\to0}{\sin x\over \cos x + 1}={\sin 0\over \cos 0+1}= {0\over 2} = 0\text{.} \end{equation*}


\begin{equation*} \lim_{x\to0}{\cos x - 1\over x}=0\text{.} \end{equation*}
Example 3.50. Limit of Other Trig Functions.

Compute the following limit \(\ds\lim_{x\to 0}\frac{\sin 5x\cos x}{x}\text{.}\)


We have

\begin{equation*} \begin{array}{rcl} \ds{\lim_{x\to 0}\frac{\sin 5x\cos x}{x}} \amp = \amp \ds{\lim_{x\to 0}\frac{5\sin 5x\cos x}{5x}}\\ \\ ~ \amp = \amp \ds{\lim_{x\to 0} 5\cos x\left(\frac{\sin 5x}{5x}\right)}\\ \\ ~ \amp = \amp 5\cdot (1)\cdot (1)~~=~~5 \end{array} \end{equation*}

since \(\cos(0)=1\) and \(\ds{\lim_{x\to 0}\frac{\sin 5x}{5x}=1}\text{.}\)

Let's do a harder one now.

Example 3.51. Limit of Other Trig Functions.

Compute the following limit: \(\ds\lim_{x\to 0}\frac{\tan^3 2x}{x^2\sin 7x}\text{.}\)


Recall that the \(\tan^3(2x)\) means that \(\tan(2x)\) is being raised to the third power.

\begin{equation*} \begin{array}{rcll} \ds{\lim_{x\to 0}\frac{\tan^3(2x)}{x^2\sin(7x)}}\amp =\amp \ds{\lim_{x\to 0}\frac{(\sin(2x))^3}{x^2\sin(7x)\cos^3(2x)}} \amp \mbox{Rewrite in terms of \(\sin\) and \(\cos\)} \\ \\ ~\amp =\amp \ds{\lim_{x\to 0}\frac{(2x)^3\left(\frac{\sin(2x)}{2x}\right)^3}{x^2(7x)\left(\frac{\sin(7x)}{7x}\right)\cos^3(2x)}} \amp \mbox{Make sine terms look like: \(\ds{\frac{\sin\theta}{\theta}}\)} \\ \\ ~\amp =\amp \ds{\lim_{x\to 0}\frac{8x^3(1)^3}{7x^3(1)(1^3)}} \amp \mbox{Replace \(\ds{\lim_{x\to 0}\frac{\sin nx}{nx}}\) with \(1\). Also, \(\cos(0)=1\).} \\ \\ ~\amp =\amp \ds{\lim_{x\to 0}\frac{8}{7}}\amp \mbox{Cancel \(x^3\)'s.} \\ \\ ~\amp =\amp \ds{\frac{8}{7}}. \end{array} \end{equation*}
Example 3.52. Applying the Squeeze Theorem.

Compute the following limit: \(\ds\lim_{x\to 0^+}x^3\cos\left(\frac{1}{\sqrt{x}}\right)\text{.}\)


We use the Squeeze Theorem to evaluate this limit. We know that \(\cos\alpha\) satisfies \(-1\leq\cos\alpha\leq 1\) for any choice of \(\alpha\text{.}\) Therefore we can write:

\begin{equation*} -1\leq\cos\left(\frac{1}{\sqrt{x}}\right)\leq 1 \end{equation*}

Since \(x\to 0^+\) implies \(x>0\text{,}\) multiplying by \(x^3\) gives:

\begin{equation*} -x^3\leq x^3\cos\left(\frac{1}{\sqrt{x}}\right)\leq x^3\text{.} \end{equation*}
\begin{equation*} \lim_{x\to 0^+}(-x^3)\leq \lim_{x\to 0^+}\left(x^3\cos\left(\frac{1}{\sqrt{x}}\right)\right)\leq\lim_{x\to 0^+} x^3\text{.} \end{equation*}

But using our rules we know that

\begin{equation*} \lim_{x\to 0^+}(-x^3)=0,\qquad\qquad \lim_{x\to 0^+} x^3=0 \end{equation*}

and the Squeeze Theorem says that the only way this can happen is if

\begin{equation*} \lim_{x\to 0^+}x^3\cos\left(\frac{1}{\sqrt{x}}\right)=0\text{.} \end{equation*}

When solving problems using the Squeeze Theorem it is also helpful to have the following theorem.

Exercises for Section 3.6.

Compute the following limits.

  1. \(\ds\lim_{x\to 0} {\sin (5x)\over x}\)


    Recall that \(\lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1\text{.}\) We will use this fact to find the desired limit by making a suitable subsitution. Let \(w = 5x\text{.}\) Then

    \begin{equation*} \begin{split} \lim\limits_{x\to 0} \frac{\sin(5x)}{x} \amp= \lim\limits_{w\to 0} \frac{\sin(w)}{w/5} \\ \amp= \lim\limits_{w\to 0} 5\frac{\sin(w)}{w} = 5(1) = 5.\end{split} \end{equation*}
  2. \(\ds\lim_{x\to 0 } {\sin(7x)\over\sin (2x)}\)

  3. \(\ds\lim_{x\to 0 } {\cot (4x) \over\csc (3x)}\)

  4. \(\ds\lim_{x\to 0 } {\tan x\over x}\)


    First, rewrite \(\tan(x)\) as \(\frac{\sin(x)}{\cos(x)}\text{.}\) Then

    \begin{equation*} \begin{split} \lim\limits_{x \to 0} \frac{\tan(x)}{x} \amp= \lim\limits_{x \to 0} \frac{\sin(x)}{x\cos(x)}\\ \amp = \lim\limits_{x \to 0} \left(\frac{\sin(x)}{x}\frac{1}{\cos(x)}\right)\\ \amp = \lim\limits_{x \to 0} \frac{\sin(x)}{x} \lim\limits_{x \to 0}\frac{1}{\cos(x)}.\end{split} \end{equation*}

    As \(x \to 0\text{,}\) \(\cos(x) \to 1\) and \(\frac{\sin(x)}{x} \to 1\text{.}\) Since both limits exist, we get by the limit laws that

    \begin{equation*} \lim\limits_{x \to 0} \frac{\tan(x)}{x} = 1. \end{equation*}
  5. \(\ds\lim_{x\to \pi/4} {\sin x-\cos x \over\cos (2x)}\)


For all \(x\geq 0\text{,}\) \(4x-9 \leq f(x) \leq x^2 - 4x +7\text{.}\) Find \(\ds\lim_{x\to4}f(x)\text{.}\)


We are given that \(4x-9\leq f(x) \leq x^2-4x+7\) for all \(x \geq 0\text{.}\) Therefore, by the Squeeze Theorem,

\begin{align*} \lim\limits_{x\to 4} (4x-9) \amp \leq \lim\limits_{x\to 4} f(x) \leq \lim\limits_{x\to 4} (x^2 -4x+7)\\ (16-9) \amp \leq \lim\limits_{x\to 4} f(x) \leq (4^2 -16 + 7)\\ 7 \amp \leq \lim\limits_{x\to 4} f(x) \leq 7 \end{align*}

Thus, \(\lim\limits_{x\to 4} f(x) = 7\text{.}\)

For all \(x\text{,}\) \(2x \leq g(x) \leq x^4 - x^2 +2\text{.}\) Find \(\ds\lim_{x\to1}g(x)\text{.}\)


Since \(2x \leq g(x) \leq x^4 - x^2+2\text{,}\) we have

\begin{align*} \lim\limits_{x\to 1} (2x) \amp \leq \lim\limits_{x\to 1} g(x) \leq \lim\limits_{x\to 1} (x^4-x^2+2)\\ 2 \amp \leq \lim\limits_{x\to 1} g(x) \leq 2 \end{align*}

So, by the Squeeze Theorem, \(\lim\limits_{x\to 1} g(x) =2\text{.}\)

Use the Squeeze Theorem to show that \(\ds\lim_{x\to0} x^4 \cos(2/x)=0\text{.}\)


We know that \(-1 \leq \cos\left(\frac{2}{x}\right) \leq 1\text{,}\) and so \(-x^4 \leq \cos \left(\frac{2}{x}\right)\leq x^4\text{.}\) Hence, by the Squeeze Theorem,

\begin{align*} \lim\limits_{x\to 0}\left(-x^4\right) \amp \leq \lim\limits_{x\to 0} \cos \left(\frac{2}{x}\right) \leq \lim\limits_{x\to 0} x^4\\ 0 \amp \leq \lim\limits_{x\to 0} \cos \left(\frac{2}{x}\right) \leq 0 \end{align*}

So we conclude that \(\lim\limits_{x\to 0} \cos \left(\frac{2}{x}\right) = 0\text{.}\)

Find the value of \(\lim\limits_{x\to\infty}\dfrac{3x+\sin x}{x+\cos x}\text{.}\) Justify your steps carefully.


We know that

\begin{equation*} 3x-1 \leq 3x + \sin x \leq 3x+1 \end{equation*}

for all \(x\text{,}\) and

\begin{equation*} x-1 \leq x+\cos x \leq 1+x \end{equation*}


\begin{equation*} \dfrac{3x-1}{x+1} \leq \dfrac{3x+\sin x}{x+\cos x} \leq \dfrac{3x+1}{x-1} \end{equation*}

so by the Squeeze Theorem,

\begin{align*} \lim\limits_{x\to \infty} \dfrac{3x-1}{x+1} \amp \leq \lim\limits_{x\to \infty} \dfrac{3x + \sin x}{x+ \cos x} \leq \lim\limits_{x\to \infty} \dfrac{3x+1}{x-1}\\ \lim\limits_{x\to \infty} \dfrac{3-\frac{1}{x}}{1+\frac{1}{x}} \amp \leq \lim\limits_{x\to \infty} \dfrac{3x + \sin x}{x+ \cos x} \leq \lim\limits_{x\to \infty} \dfrac{3+\frac{1}{x}}{1-\frac{1}{x}}\\ 3 \amp \leq \lim\limits_{x\to \infty} \dfrac{3x+\sin x}{x+\cos x} \leq 3 \end{align*}

Thus, \(\lim\limits_{x\to \infty} \dfrac{3x+\sin x}{x+\cos x} = 3\text{.}\)