1.3: Squeeze Theorem and Trigonometric Limits

In Section 1.2 we learned techniques to find limits analytically. In this section we extend these ideas as we evaluate limits of trigonometric functions and composite functions. We discuss the following topics:

• Squeeze Theorem
• Trigonometric Limits
• Limits by Change of Variable

Squeeze Theorem

Suppose that a function \(f\) is bounded between two other functions \(g\) and \(h.\) Mathematically, on some interval that contains the number \(a,\) let \(g(x) \leq f(x) \leq h(x)\) (except possibly at \(a\) itself). If \(\lim_{x \to a} g(x)\) and \(\lim_{x \to a} h(x)\) both equal \(L,\) then the Squeeze Theorem asserts that \(\lim_{x \to a} f(x) = L\) as well. (The Squeeze Theorem is also called the Sandwich Theorem or Pinching Theorem.) As shown by Figure 1, as \(x\) is made closer and closer to \(a,\) the graphs of \(g\) and \(h\) trap the graph of \(f,\) thus causing the value of \(f(x)\) to be pinched toward \(L.\)

Proving the Squeeze Theorem requires the formal definition of a limit (Section 1.5), so its proof is presented in Appendix A.2.

It's wrong to say \[\lim_{x \to 0} x^2 \sin \par{\frac{1}{x}} = \par{\lim_{x \to 0} x^2} \par{\lim_{x \to 0} \sin \par{\frac{1}{x}}}\] because \(\lim_{x \to 0} \sin(1/x)\) does not exist (Example 1.1-5). The Product Law is only applicable if both limits exist. Instead, let's bound \(x^2 \sin(1/x)\) between two functions that have the same limit as \(x \to 0.\) Since the range of sine is \([-1, 1],\) we have \[-1 \leq \sin \par{\frac{1}{x}} \leq 1 \pd\] Multiplying every term by \(x^2\) shows \[ -x^2 \leq x^2 \sin \par{\frac{1}{x}} \leq x^2 \pd \] Observe that \[\lim_{x \to 0} \par{-x^2} = \lim_{x \to 0} \par{x^2} = 0 \pd\] Hence, the Squeeze Theorem gives \[\lim_{x \to 0} x^2 \sin \par{\frac{1}{x}} = \boxed 0\] (See Figure 2.)

Trigonometric Limits

Just as with limits of polynomials and rational expressions, limits of trigonometric functions can be evaluated by Direct Substitution whenever they are defined. We therefore attain the following limits.

In addition to these rules, trigonometric identities help us evaluate limits of trigonometric functions. The most important one is the Pythagorean identity: \begin{equation} \sin^2 x + \cos^2 x = 1 \pd \label{eq:pythag-identity} \end{equation} We also use the double-angle identities for sine and cosine, given as follows: \begin{align} \sin 2x &= 2 \sin x \cos x \label{eq:sin-double-angle} \cma \nl \cos 2x &= \cos^2 x - \sin^2 x \label{eq:cos-double-angle} \pd \end{align}

A Special Limit An important limit for working with trigonometric functions is \begin{equation} \lim_{x \to 0} \frac{\sin x}{x} = 1 \cma \label{eq:sin-lim-1} \end{equation} where \(x\) is in radians. To prove it, we present a geometric argument that concludes with the Squeeze Theorem. Let's begin by drawing a Unit Circle and noting the central angle \(\theta,\) as shown by Figure 3. The inscribed triangle \(\Delta OPQ\) has width \(1\) and height \(\sin \theta;\) its area is therefore \(\tfrac{1}{2} \sin \theta.\) Similarly, triangle \(\Delta ORQ\) has width \(1\) and height \(\length{RQ} = \tan \theta\) (since \(\tan \theta = \length{RQ}/1\)), so its area is \(\tfrac{1}{2} \tan \theta.\) Conversely, the circular sector with arc \(\arcHat{PQ}\) has area \(\tfrac{1}{2} (1)^2 \theta\) \(= \tfrac{1}{2} \theta.\) By comparing the overlapping areas, we see \[ \underbrace{\tfrac{1}{2} \tan \theta}_{\text{area of } \Delta ORQ} \geq \underbrace{\tfrac{1}{2} \theta}_{\substack{\text{area of sector} \\ \text{with arc } \arcHat{PQ}}} \geq \underbrace{\tfrac{1}{2} \sin \theta}_{\text{area of } \Delta OPQ} \pd \] Note that each quantity is positive because \(0 \lt \theta \lt \pi/2.\) Let's multiply each expression by \(2/\sin \theta;\) doing so gives \[ \frac{1}{\cos \theta} \geq \frac{\theta}{\sin \theta} \geq 1 \pd \] We now take the reciprocal of each fraction; in doing so, we must also flip the directions of the inequality signs because each quantity is positive. (For example, \(3 \gt 2 \iffArrow\) \(\tfrac{1}{3} \lt \tfrac{1}{2}.\)) Doing so gives \begin{equation} \cos \theta \leq \frac{\sin \theta}{\theta} \leq 1 \pd \label{eq:sin-lim-inequality} \end{equation} Because cosine is an even function, we have \(\cos \theta = \cos(-\theta).\) Likewise, since sine is an odd function, we get \((\sin \theta)/\theta\) \(= [-\sin(-\theta)]/(-\theta).\) Accordingly, the inequality in \(\eqref{eq:sin-lim-inequality}\) is valid for all nonzero \(\theta\) in \((-\pi/2, \pi/2).\) Note that \[\underbrace{\lim_{\theta \to 0} \cos \theta}_{\cos 0} = \lim_{\theta \to 0} 1 = 1 \pd\] Hence, the Squeeze Theorem gives \[\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \cma\] thus proving \(\eqref{eq:sin-lim-1}.\)

Let's summarize these special trigonometric limits for easy reference. We use these limits to resolve many indeterminate limits that contain trigonometric functions. Thus, it is important to have them memorized. (We will then apply these limits in Section 2.2 as we discuss derivatives.)

Limits by Change of Variable

When we have the limit of a complicated function, it may be useful to perform a change of variable to attain a simpler function. Changing the variable is particularly useful when we compute the limit of a composite function. Mathematically, we use the following fact.