OBJECTIVES

When you finish this section, you should be able to:

1. Use the Squeeze Theorem to find a limit (p. 107)
2. Find limits involving trigonometric functions (p. 108)
3. Determine where the trigonometric functions are continuous (p. 111)
4. Determine where an exponential or a logarithmic function is continuous (p. 113)

THEOREM Squeeze Theorem

Suppose the functions \(f\), \(g\), and \(h\) have the property that for all \(x\) in an open interval containing \(c\), except possibly at \(c\), \begin{equation*} f(x)\leq g(x)\leq h(x)\ \end{equation*}

If \begin{equation*} \lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}h(x)=L\ \end{equation*}

then \begin{equation*} \lim_{x\rightarrow c}g(x)=L\ \end{equation*}

Using the Squeeze Theorem to Find a Limit

Use the Squeeze Theorem to find \(\lim\limits_{x\rightarrow 0}\left( x\sin \dfrac{1}{x}\right)\).

Solution If \(x\neq 0\), then \(\sin \dfrac{1}{x}\) is defined. We seek two functions that “squeeze” \(y=x\sin \dfrac{1}{x}\) near \(0.\) Since \(-1\leq \sin x\leq 1\) for all \(x,\) we begin with the inequality \begin{equation*} \left\vert \sin \dfrac{1}{x}\right\vert \leq 1\qquad x\neq 0 \end{equation*}

Since \(x\neq 0\) and we seek to squeeze \(x\sin \dfrac{1}{x},\) we multiply both sides of the inequality by \(\vert x\vert ,\) \(x\neq 0.\) Since \(\vert x\vert >0,\) the direction of the inequality is preserved. Note that if we multiply \(\left\vert \sin \dfrac{1}{x}\right\vert \leq 1\) by \(x,\) we would not know whether the inequality symbol would remain the same or be reversed since we do not know whether \(x>0\) or \(x\lt 0.\) \begin{equation*} \begin{array}{r@{\qquad}ll} |x|\left\vert \sin \dfrac{1}{x}\right\vert \leq |x| & & {\color{#0066A7}{\hbox{Multiply both sides by \(\vert x \vert >0\).}}} \\[12pt] \left\vert x\sin \dfrac{1}{x}\right\vert \leq |x| & & {\color{#0066A7}{{|a|\cdot |b| = |a\,b|}.}} \\[11pt] -\vert x\vert \leq x\sin \dfrac{1}{x}\leq |x| & & {\color{#0066A7}{\vert a\vert\,{\leq}\,b\hbox{ is equivalent to } {-b\leq a\leq b.}}} \end{array} \end{equation*}

Figure 39 \(g(x)=x\sin\dfrac{1}{x}\) is squeezed between \(f(x)=-\vert x\vert\) and \(h(x)=\vert x\vert\).

Now use the Squeeze Theorem with \(f(x) = - |x|\), \(g(x) = x\sin \dfrac{1}{x}\), and \(h(x) = |x|\). Since \(f(x)\leq g(x)\leq h(x)\) and \begin{equation*} \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}(-|x|)=0\qquad \hbox{and}\qquad \lim_{x\rightarrow 0}h(x)=\lim_{x\rightarrow 0}\vert x\vert =0 \end{equation*}

it follows that \[ \lim_{x\rightarrow 0}g(x)=\lim_{x\rightarrow 0}\left( x\cdot \sin \frac{1}{x} \right) =0 \]

NOW WORK

Problem 5.

THEOREM Two Basic Trigonometric Limits

\[\bbox[5px, border:1px solid black, #F9F7ED]{\lim\limits_{x \rightarrow 0}\sin x=0 \qquad \lim\limits_{x\rightarrow 0}\cos x=1} \]

THEOREM

If \(\theta\) is measured in radians, then \[\bbox[5px, border:1px solid black, #F9F7ED]{ \lim\limits_{\theta \rightarrow 0}\dfrac{\sin \theta }{\theta }=1} \]

Proof

Although \(\dfrac{\sin \theta}{\theta}\) is a quotient, we have no way to divide out \(\theta\). To find \(\lim\limits_{\theta \rightarrow 0^{+}}\dfrac{\sin \theta}{\theta}\), we let \(\theta \) be a positive acute central angle of a unit circle, as shown in Figure 42(a). Notice that \({\it COP}\) is a sector of the circle. We add the point \(B = (\cos \theta ,0)\) to the graph and form triangle BOP. Next we extend the terminal side of angle \(\theta \) until it intersects the line \(x = 1\) at the point D, forming a second triangle COD. The \(x\)-coordinate of D is \(1\). Since the length of the line segment \(\overline{OC}\) is \({OC} = 1\), then the \(y\)-coordinate of D is \({CD} = \dfrac{{CD}}{{OC}} = \tan \theta\). So, \(D = (1,\tan \theta)\). See Figure 42(b).

Figure 42

We see from Figure 42(b) that

area of triangle \(BOP\)≤ area of sector \(COP\)≤ area of triangle COD \(\tag{1}\)

Each of these areas can be expressed in terms of \(\theta \) as \[ \begin{array}{l@{\quad}l} \hbox{area of triangle}{\it BOP} = \dfrac{1}{2}\cos \theta \cdot \sin \theta & {\color{#0066A7}{\hbox{area of a triangle $ = \dfrac{1}{2}$ base $\times$ height}}}\\[6pt] \hbox{area of sector}{\it COP} = \dfrac{\theta}{2} & {\color{#0066A7}{\hbox{area of sector $A = \dfrac{1}{2}{r}^{2}{\theta}$; $r = 1$}}}\\[6pt] \hbox{area of triangle}{\it COD} = \dfrac{1}{2}\cdot 1\cdot \tan \theta = \dfrac{ \tan \theta}{2} & {\color{#0066A7}{\hbox{area of a triangle $ = \dfrac{1}{2}$ base $\times$ height}}} \end{array} \]

So, for \(0\lt \theta \lt \dfrac{\pi}{2}\), we have \begin{equation*} \begin{array}{@{\;}rcl@{\quad}l} \!\!\dfrac{1}{2}\cos \theta \cdot \sin \theta &\leq& \dfrac{\theta}{2}\leq \dfrac{ \tan \theta}{2} & {\color{#0066A7}{\hbox{(1)}}} \\[10pt] &&&\quad \\ \cos \theta \cdot \sin \theta &\leq& \theta \leq \dfrac{\sin \,\theta}{\cos \theta} & {\color{#0066A7}{\hbox{Multiply all parts by 2; $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$.}}}\\[11pt] \cos \theta &\leq& \dfrac{\theta}{\sin \theta}\leq \dfrac{1}{\cos \theta} & {\color{#0066A7}{\hbox{Divide all parts by $\sin \theta$; since $0 \lt \theta \lt \dfrac{\pi}{2}, \sin \theta >0$.}}} \\[10pt] \dfrac{1}{\cos \theta}&\geq& \dfrac{\sin \theta}{ \theta}\geq \cos \theta & {\color{#0066A7}{\hbox{Invert each term in the inequality; change the inequality signs.}}}\\[-2.5pt] \cos \theta &\leq& \dfrac{\sin \theta}{\theta} \leq \dfrac{1}{\cos \theta} & {\color{#0066A7}{\hbox{Rewrite using $\leq$.}}} \end{array} \end{equation*}

Now we use the Squeeze Theorem with \(f(\theta) = \cos \theta\), \(\ g(\theta) = \dfrac{\sin \theta}{\theta},\) and \(h(\theta) = \dfrac{1}{\cos \theta}\). Since \(f(\theta)\leq \,g(\theta)\leq h(\theta)\), and since \begin{equation*} \lim_{\theta \rightarrow 0^{+}}f(\theta) = \lim_{\theta \rightarrow 0^{+}}\cos \theta = 1\quad \hbox{and}\quad \lim_{\theta \rightarrow 0^{+}}h(\theta) = \lim_{\theta \rightarrow 0^{+}}\frac{1}{\cos \theta} = \frac{\lim\limits_{\theta \rightarrow 0^{+}}1}{\lim\limits_{\theta \rightarrow 0^{+}}\cos \theta} = \frac{1}{1} = 1 \end{equation*} then by the Squeeze Theorem, \(\lim\limits_{\theta \rightarrow 0^{+}}g(\theta) = \lim\limits_{\theta \rightarrow 0^{+}}\dfrac{\sin \theta}{\theta} = 1.\)

Now we use the fact that \(g(\theta) = \dfrac{\sin \theta}{\theta}\) is an even function, that is, \[ g(-\theta) = \dfrac{\sin \left(-\theta \right)}{-\theta} = \dfrac{ -\sin \theta}{-\theta} = \dfrac{\sin \theta}{\theta} = g(\theta) \]

So, \begin{equation*} \lim\limits_{\theta \rightarrow 0^{-}}\dfrac{\sin \theta}{\theta} = \lim\limits_{\theta \rightarrow 0^{+}}\dfrac{\sin \left(-\theta \right) }{-\theta} = \lim\limits_{\theta \rightarrow 0^{+}}\dfrac{\sin \theta}{ \theta} = 1 \end{equation*}

It follows that \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin \theta}{\theta} = 1\).

Finding the Limit of a Trigonometric Function

Find:

1. \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (3\theta)}{\theta}\)
2. \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (5\theta)}{\sin (2\theta)}\)

Solution (a) Since \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (3\theta)}{\theta}\) is not in the same form as \(\lim\limits_{\theta\rightarrow 0}\dfrac{\sin \theta}{\theta}\), we multiply the numerator and the denominator by 3, and make the substitution \(t = 3\theta\). \[ \begin{equation*} \lim_{\theta \rightarrow 0}\frac{\sin (3\theta )}{\theta }=\lim_{\theta \rightarrow 0}\frac{3\sin (3\theta )}{3\theta } \underset{\color{#0066A7}{{t\rightarrow 0}{\hbox{ as }}{\theta \rightarrow 0}}}{\underset{\color{#0066A7}{{t=3\theta}}}{\underset{\color{#0066A7}{\uparrow }}{=}}} \lim_{t\rightarrow 0}\left( 3\frac{\sin t}{t}\right) =3\lim_{t\rightarrow 0} \frac{\sin t}{t}\underset{\color{#0066A7}{{\lim\limits_{t\rightarrow 0}\dfrac{\sin t}{t}=1 }}}{\underset{\color{#0066A7}{\uparrow }}{=}}\!\!(3)(1)=3 \end{equation*} \]

(b) We begin by dividing the numerator and the denominator by \(\theta\). Then \[ \begin{equation*} \dfrac{\sin (5\theta)}{\sin (2\theta)} = \dfrac{ \dfrac{\sin (5\theta)}{\theta}}{\dfrac{\sin (2\theta)}{\theta}} \end{equation*} \]

Now we follow the approach in (a) on the numerator and on the denominator. \[ \begin{eqnarray*} \lim\limits_{\theta \rightarrow 0}\dfrac{\sin ( 5\theta ) }{ \theta } &=&\lim\limits_{\theta \rightarrow 0}\dfrac{5\sin ( 5\theta ) }{5\,\theta } \underset{\color{#0066A7}{{t\rightarrow 0\hbox{ as }\theta \rightarrow 0}}} {\underset{\color{#0066A7}{t=5\theta}}{\underset{\color{#0066A7}{\uparrow }}{=}}}\lim\limits_{t\rightarrow 0} \dfrac{5\sin t}{t}=5\lim\limits_{t\rightarrow 0}\left( \dfrac{\sin t}{t} \right) =5 \\ \lim\limits_{\theta \rightarrow 0}\dfrac{\sin ( 2\theta ) }{ \theta } &=&\lim\limits_{\theta \rightarrow 0}\dfrac{2\sin ( 2\theta ) }{2\theta } \underset{\color{#0066A7}{{t\rightarrow 0\hbox{ as }\theta \rightarrow 0}}}{ \underset{\color{#0066A7}{{t=2\theta}}}{\underset{\uparrow }{=}}}\lim\limits_{t\rightarrow 0}\left( \dfrac{2\sin t}{t}\right) =2\lim\limits_{t\rightarrow 0}\dfrac{\sin t}{t}=2 \\ \lim\limits_{\theta \rightarrow 0} \dfrac{\sin (5\theta)}{\sin (2\theta)} &=&\dfrac{ \lim\limits_{\theta \rightarrow 0}\dfrac{\sin (5\theta)}{\theta}}{\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (2\theta)}{\theta}} = \frac {5}{2} \end{eqnarray*} \]

NOW WORK

Problems 23 and 25.

Finding a Basic Trigonometric Limit

Establish the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{ \lim\limits_{\theta \rightarrow 0}\dfrac{\cos \theta -1}{\theta} = 0} \]

where \(\theta\) is measured in radians.

Solution First we rewrite the expression \(\dfrac{\cos \theta -1}{\theta}\) as the product of two terms whose limits are known. For \(\theta \neq 0\), \[ \begin{eqnarray*} \frac{\cos \theta -1}{\theta }&=&\left( \frac{\cos \theta -1}{\theta }\right)\! \left( \frac{\cos \theta +1}{\cos \theta +1}\right)\\ &=&\frac{\cos ^{2}\theta -1 }{\theta (\cos \theta +1)}\underset{\color{#0066A7}{{\sin ^{2}{\theta +}\cos ^{2} {\theta =1}}}}{\underset{\color{#0066A7}{\uparrow }}{=}}\frac{-\sin ^{2}\theta }{\theta (\cos \theta +1)} = \left( \frac{\sin \theta }{\theta }\right) \frac{(-\sin \theta)}{\cos \theta +1} \end{eqnarray*} \]

Now we find the limit. \[ \begin{eqnarray*} \lim_{\theta \rightarrow 0}\frac{\cos \theta -1}{\theta } &=&\lim_{\theta \rightarrow 0}\left[ \left( \frac{\sin \theta }{\theta } \right) \!\left( \frac{(-\sin \theta )}{\cos \theta +1}\right) \right] = \left[ \lim_{\theta \rightarrow 0}\frac{ \sin \theta }{\theta }\right]~ \!\left[ \lim_{\theta \rightarrow 0} \frac{ -\sin\theta}{\cos \theta +1} \right] \nonumber \\ &=& 1\,{\cdot}\,\dfrac{\lim\limits_{\theta \rightarrow 0}(-\sin \theta)}{\lim\limits_{\theta \rightarrow 0}(\cos\theta+1)} =\dfrac{0}{2} =0 \end{eqnarray*} \]

NOW WORK

Problem 31.

Showing \(f(x)\) = sin \(x\) Is Continuous at 0

• \(f(0) = \sin 0 = 0\), so \(f\) is defined at 0.
• \(\lim\limits_{x\rightarrow 0}f(x) = \lim\limits_{x\rightarrow 0}\sin x = 0\), so the limit at 0 exists.
• \(\lim\limits_{x\rightarrow 0}\sin x = \sin 0 = 0\).

Since all three conditions of continuity are satisfied, \(f(x) = \sin x\) is continuous at 0.

THEOREM

• The sine function \(y = \sin x\) is continuous on its domain, all real numbers.
• The cosine function \(y = \cos x\) is continuous on its domain, all real numbers.

Application: Projectile Motion

Figure 43

An object is propelled from ground level at an angle \(\theta \), \(0\lt \theta \lt \dfrac{\pi}{2}\), to the horizontal with an initial velocity of 16feet/second, as shown in Figure 43. The equations of the horizontal position \(x = x(\theta)\) and the vertical position \(y = y(\theta)\) of the object after t seconds are given by \begin{equation*} x = x(\theta) = (16\cos \theta) t\qquad \hbox{and}\qquad y = y(\theta) = -16t^{2}+(16\sin \theta) t \end{equation*}

For a fixed time t:

1. Find \(\lim\limits_{\theta \rightarrow 0^{+}}x(\theta)\) and \(\lim\limits_{\theta \rightarrow 0^{+}}y(\theta)\).
2. Are the limits found in (a) consistent with what is expected physically?
3. Find \(\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}x(\theta)\) and \(\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}y(\theta)\).
4. Are the limits found in (c) consistent with what is expected physically?

Solution (a) \begin{eqnarray*} \lim\limits_{\theta \rightarrow 0^{+}}x(\theta) &=&\Big[\lim\limits_{\theta \rightarrow 0^{+}}(16\cos \theta)\Big] t = 16t\qquad{\color{#0066A7}{\hbox{$t$ fixed}}}\\[3pt] \lim\limits_{\theta \rightarrow 0^{+}}y(\theta) &=&\lim\limits_{\theta \rightarrow 0^{+}} [ -16t^{2}+(16\sin \theta) t ]\\[3pt] &=&\lim\limits_{\theta \rightarrow 0^{+}}(-16t^{2}) +\Big[\lim\limits_{\theta \rightarrow 0^{+}}(16\sin \theta)\Big] t = -16t^{2} \end{eqnarray*}

(b) The limits in (a) tell us that the horizontal position x of the object moves to the right (\(x = 16t\)) over time and that the vertical position y of the object immediately moves downward (\(y = -16t^{2}\)) into the ground. Neither of these conclusions is consistent with what we expect from the motion of an object propelled at an angle close to \(\theta = 0\).

(c) \begin{eqnarray*} \lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}x(\theta) &=&\Big[\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}(16\cos \theta)\Big] t = 0\\[3pt] \lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}y(\theta) &=&\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}\left[ -16t^{2}+(16\sin \theta) t\right] = \lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}(-16t^{2}) +\Big[\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}(16\sin \theta)\Big] t\\[3pt] &=&-16t^{2}+16t \end{eqnarray*}

(d) The limits we found in (c) tell us that the horizontal position remains at \(0\). The vertical position \(y = -16t^{2}+16t = -16t(t-1)\) tells us the object has a maximum height of 4feet and returns to the ground after 1 second. (The parabola \(y = -16t^{2}+16t\) opens down and has a maximum value at \(t = \dfrac{-b}{2a} = \dfrac{-16}{-32} = \dfrac{1}{2}\) at which time \(y = 4\).) These conclusions are consistent with what we would expect from the motion of an object propelled close to the vertical.

NOW WORK

Problem 55.

• \(y = \tan x\): Since \(\tan x = \dfrac{\sin x}{\cos x}\), from the Continuity of a Quotient property, \(y = \tan x\) is continuous at all real numbers except those for which \(\cos x = 0\). That is, it is continuous on its domain, all real numbers except odd multiples of \(\dfrac{\pi}{2}\).
• \(y = \sec x\): Since \(\sec x = \dfrac{1}{\cos x}\), from the Continuity of a Quotient property, \(y = \sec x\) is continuous at all real numbers except those for which \(\cos x = 0\). That is, it is continuous on its domain, all real numbers except odd multiples of \(\dfrac{\pi}{2}\).

• \(y = \cot x\): Since \(\cot x = \dfrac{\cos x}{\sin x}\), from the Continuity of a Quotient property, \(y = \cot x\) is continuous at all real numbers except those for which \(\sin x = 0.\)

  113

  That is, it is continuous on its domain, all real numbers except integer multiples of \(\pi\).

• \(y = \csc x\): Since \(\csc x = \dfrac{1}{\sin x},\) from the Continuity of a Quotient property, \(y = \csc x\) is continuous at all real numbers except those for which \(\sin x = 0.\) That is, it is continuous on its domain, all real numbers except integer multiples of \(\pi \).

NEED TO REVIEW?

Inverse trigonometric functions are discussed in Section P.7, pp. 58-63.

THEOREM Continuity of Exponential and Logarithmic Functions

• An exponential function is continuous on its domain.
• A logarithmic function is continuous on its domain.

Show that a Function Is Continuous

Show that:

1. \(f(x) = e^{2x}\) is continuous for all real numbers.
2. \(F(x) = \sqrt[3]{\ln x}\) is continuous for \(x>0\).

Solution (a) The domain of the exponential function is the set of all real numbers, so \(f\) is defined for any number \(c\). That is, \(f(c) = e^{2c}\). Also for any number \(c\), \begin{equation*} \lim\limits_{x\rightarrow c}f(x) = \lim\limits_{x\rightarrow c}e^{2x} = \lim\limits_{x\rightarrow c}(e^{x}) ^{2} = \left[ \lim\limits_{\kern.5ptx\rightarrow c}e^{x}\right] ^{2} = (e^{c}) ^{2} = e^{2c} = f(c) \end{equation*}

Since \(\lim\limits_{x\rightarrow c}f(x) = f(c)\) for any number \(c,\) then \(f\) is continuous at all numbers \(c\).

(b) The logarithmic function \(f(x) = \ln x\) is continuous on its domain, the set of all positive real numbers. The function \(g(x) = \sqrt[3]{x}\) is continuous on its domain, the set of all real numbers. Then for any real number \(c>0\), the composite function \(F(x) = (g\circ f) (x) = \) \(\sqrt[3]{\ln x}\) is continuous at \(c\). That is, \(F\) is continuous at all numbers \(x>0\).

NOW WORK

Problem 45.