4.2 Limits and Continuity of Trigonometric Functions

4.2.1 Continuity of Trigonometric Functions

 

We use the following theorem, which we state without proof, to prove that sine and cosine are continuous on \(\mathbb{R}\).

Theorem 1. (The Squeeze Theorem.) Assume that \(f\), \(g\), and \(h\) are defined on an interval \(I\) containing \(a\) (except possibly at \(a\)), and assume further that
\[f(x)\leq g(x)\leq h(x) \;\; \mathrm{for \; all} \;\; x \in I-\{a\}, \;\;\; \mathrm{and} \;\;\;
\lim_{x\to a}f(x)=L=\lim_{x\to a}h(x).\]
Then, it is also true that \(\displaystyle \lim_{x\to a}g(x)=L\).

 

Proposition 1. (a) \(\sin x\) is continuous at \(x=0\);

(b) \(\cos x\) is continuous at \(x=0\).

Proof. Let \(\displaystyle x \in \left(0,\frac{\pi}{2}\right)\), then, in the notation of Figure 4.1, where the circle is the unit circle, we have
\[\mathrm{Area}(\triangle OAB) =\frac{1}{2}(1)(\sin x)\;\;\; \mathrm{and} \;\;\; \mathrm{Area}(\mathrm{circular\; sector\;} OAB)=\frac{1}{2}(1)^2x.\]
Furthermore
\[\mathrm{Area}(\triangle OAB)\leq \mathrm{Area}(\mathrm{circular\; sector\;} OAB)\;\; \Longrightarrow \;\; \frac{1}{2}(1)(\sin x)\leq \frac{1}{2}(1)^2x.\]
Thus,
\begin{equation}\label{ch04-02-eq01}
0 < \sin x\leq x\;\; \text{for all}\;\; x\in \left (0,\frac{\pi}{2}\right ]\tag{4.1}. \end{equation} (a) To prove that sine is continuous at \(x=0\), we need to verify three things:

(i) \(0 \in D_{\sin}\),

(ii) \(\displaystyle \lim_{x\to 0}\sin x \) exists, and

(iii) \(\displaystyle \lim_{x\to 0}\sin x =\sin 0\).

Part (i) follows from the fact that \(D_{\sin}=\mathbb{R}\). Furthermore \(\sin 0 =0\).

Parts (ii) and (iii) follow, for \(\displaystyle 0 < x < \frac{\pi}{2}\), from the Squeeze Theorem, and (\ref{ch04-02-eq01}), since \(\displaystyle \lim_{x\to 0} x = 0\). Thus, sine is continuous from the right at \(x=0\).

It remains to prove that sine is continuous from the left at \(x=0\). For this, note that since sine is an odd function, \(\sin x = – \sin (-x)\). Thus, if \(x<0\), then \(-x>0\). It follows that sine is also continuous from the left. Hence, sine is continuous at \(x=0\).

(b) To prove that cosine is continuous at \(x=0\), we need to verify three things:

(i) \(0 \in D_{\cos}\), but this is true because \(D_{\cos}\) and \(\cos 0 =1\).

(ii) \(\displaystyle \lim_{x\to 0}\cos x \) exists, and

(iii) \(\displaystyle \lim_{x\to 0}\cos x =\cos 0\).

Note that parts (ii) and (iii) are equivalent to prove that \(\displaystyle \lim_{x\to 0}(1-\cos x)=0\).

Since
\[0\leq 1-\cos x<(1-\cos x) (1 + \cos x)= 1-\cos^2x=\sin^2x, x\in \left [-\frac{\pi}{2},\frac{\pi}{2}\right ],\] and since \[\lim_{x\to 0}\sin^2 x = \left(\lim_{x\to 0}\sin x\right)\left(\lim_{x\to 0}\sin x\right) =0\cdot 0=0,\] by the Squeeze Theorem, (ii) and (iii) follow. Thus, cosine is continuous at \(x=0\).

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Theorem 2. \(\sin x\), \(\cos x\), \(\tan x\), \(\cot x\), \(\sec x\) and \(\csc x\) are continuous everywhere in their domains.

Proof.
To prove that sine is continuous at \(x=a\), for any \(a\in \mathbb{R}\), we need to prove that

(i) \(a \in D_{\sin}\), but this is true because \(D_{\sin}=\mathbb{R}\), and

(ii) and (iii) \(\displaystyle \lim_{x\to a}\sin x =\sin a\).

Set \(h=x-a\), so that \(x=a+h\), and \(x\to a \Longleftrightarrow h\to 0\). Thus, we only need to prove that
\[\lim_{h\to 0}\sin (a+h) =\sin a.\]
We use the addition formula:
\begin{align*}
\lim_{h\to 0}\sin (a+h)& =\lim_{h\to 0}(\sin a \cos h + \sin h \cos a )
=\lim_{h\to 0} (\sin a \cos h) +\lim_{h\to 0} ( \sin h \cos a ) =\\
&=\left(\lim_{h\to 0} \sin a\right)\left(\lim_{h\to 0}\cos h\right) +\left(\lim_{h\to 0}\sin h\right)\left(\lim_{h\to 0}\cos a\right)=\\
&=(\sin a)(1)+(0)(\cos a)=\sin a
\end{align*}
Note that the above calculation is valid because all limits involved exist. Thus we can use the rules for limits.

A similar calculation proves the continuity of cosine on \(\mathbb{R}\).

The remaining trigonometric functions are continuous on their domains because they are quotients of continuous functions, and thus continuous everywhere the denominator does not vanish, that is, in their domains.

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4.2.2 Limits of Trigonometric Functions

 

We prove two important limits that will be used to compute the derivatives of the trigonometric functions.

Theorem 3.
\begin{equation}\label{ch04-02-eq02}
\lim_{x\to 0}\frac{\sin x}{x} =1\tag{4.2}.
\end{equation}

Proof.
Assume \(\displaystyle 0 < x < \frac{\pi}{2}\). On one hand, from (\ref{ch04-02-eq01}) we have \(\displaystyle 0 < \frac{\sin x}{x} < 1\). On the other hand, in the notation of Figure 4.1, we have
\(\displaystyle \mathrm{Area}(\triangle OAC)\underset{?}{=} \frac{1}{2}(1)\tan x\), and since
\(\mathrm{Area}(\mathrm{circular\; sector\;} OAB) \leq \mathrm{Area}(\triangle OAC)\), it follows that
\[\frac{1}{2}x < \frac{1}{2}\tan x =\frac{1}{2}\frac{\sin x}{\cos x}\Longleftrightarrow \cos x < \frac{\sin x}{x} \Longrightarrow \cos x < \frac{\sin x}{x} < 1 \] (note that \(x\), \(\sin x\) and \(\cos x\) are all positive). Applying the Squeeze Theorem, we obtain (\ref{ch04-02-eq02}).

Finally, for \(\displaystyle -\frac{\pi}{2} < x < 0\), note that \[\frac{\sin x}{x} =\frac{-\sin x}{-x}=\frac{\sin (-x)}{-x},\] and since now \(-x > 0\), letting \(x\to 0\) is equivalent to letting \(-x\to 0\), and the result follows.

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Corollary 1.
\begin{equation}\label{ch04-02-eq03}
\lim_{x\to 0}\frac{1-\cos x}{x} =0\tag{4.3}.
\end{equation}

Proof. Assume \(0 < x <\frac{\pi}{2}\). Then \(0< \cos x < 1\), that is, \(0 < 1- \cos x\), and \(1 < 1+\cos x\). Thus \[ 0<\frac{1-\cos x}{x} < \frac{1-\cos x}{x}(1+\cos x) =\frac{1-\cos^2 x}{x}=\frac{\sin^2 x}{x}=\frac{\sin x}{x} \sin x.\] Applying the Squeeze Theorem, (\ref{ch04-02-eq02}), and the continuity of sine the result follows. For \(-\frac{\pi}{2} < x < 0\), a similar argument yields the result.

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Example 1. Evaluate the limit.
(a) \(\displaystyle \lim_{x \to 0}\frac{\tan 2x}{x}\);      (b) \(\displaystyle \lim_{x \to 0}\frac{\tan 2x}{\sin 3x}\);     (c) \(\displaystyle \lim_{x \to 0}\frac{1-\cos x}{x^2}\);     (d) \( \displaystyle \lim_{x \to a} \frac{\sin(\sqrt[3]{x}-\sqrt[3]{a})}{x-a}\).

Solution. (a) First we need to rewrite the limit in a way that in involves either (\ref{ch04-02-eq02}) or (\ref{ch04-02-eq03}) (or both). For this, note that
\[ \frac{\tan 2x}{x} =\frac{\sin 2x}{x\cos 2 x}=\frac{2\sin 2x}{2x}\frac{1}{\cos 2 x}.\]
Note also that we multiplied and divided by 2. The idea is to introduce a new variable \(\theta= 2x\), and since \(x\to 0 \Longleftrightarrow \theta \to 0\), we have
\[\lim_{x \to 0}\frac{\tan 2x}{x}
=\lim_{x \to 0}\frac{2\sin 2x}{2x}\frac{1}{\cos 2 x}
=\lim_{\theta \to 0}\frac{2\sin \theta}{\theta}\frac{1}{\cos \theta}=2\cdot 1\cdot \frac{1}{1}=2.\]

(b)
\[\lim_{x \to 0}\frac{\tan 2x}{\sin 3x}=\lim_{x \to 0}\frac{\sin 2x}{\sin 3x}
=\lim_{x \to 0}\frac{2\sin 2x}{2x}\frac{3x}{3\sin 3x}=
\left(\lim_{x \to 0}\frac{2\sin 2x}{2x}\right)\left(\lim_{x \to 0}\frac{3x}{3\sin 3x}\right)
=\frac{2}{3}.\]

(c)
\begin{align*}
\lim_{x \to 0}\frac{1-\cos x}{x^2}
&=\lim_{x \to 0}\frac{1-\cos x}{x^2} \frac{1+\cos x}{1+\cos x}
=\lim_{x \to 0}\frac{1-\cos^2 x}{x^2} \frac{1}{1+\cos x}
=\lim_{x \to 0}\frac{\sin^2 x}{x^2} \frac{1}{1+\cos x}=\\
&=\left(\lim_{x \to 0}\frac{\sin x}{x }\right) \left(\lim_{x \to 0}\frac{\sin x}{x }\right)\left(\lim_{x \to 0} \frac{1}{1+\cos x}\right) = 1\cdot 1 \cdot \frac{1}{2}= \frac{1}{2}.
\end{align*}

(d)
\begin{align*}
\lim_{x \to a} \frac{\sin(\sqrt[3]{x}-\sqrt[3]{a})}{x-a}
&=\lim_{x \to a} \frac{\sin(\sqrt[3]{x}-\sqrt[3]{a})}{(\sqrt[3]{x})^3-(\sqrt[3]{a})^3}
=\lim_{x \to a} \frac{\sin(\sqrt[3]{x}-\sqrt[3]{a})}{(\sqrt[3]{x}-\sqrt[3]{a})((\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{a}+(\sqrt[3]{a})^2)} =\\
& = \left(\lim_{x \to a} \frac{\sin(\sqrt[3]{x}-\sqrt[3]{a})}{\sqrt[3]{x}-\sqrt[3]{a}}\right)
\left(\lim_{x \to a} \frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}\sqrt[3]{a}+(\sqrt[3]{a^2}}\right) = \frac{1}{3\sqrt[3]{a^2}},
\end{align*}
where, for the limit in the first factor, we introduced the substitution \(\theta =\sqrt[3]{x}-\sqrt[3]{a}\), while, for the limit in the second factor, we made a direct substitution using the continuity of \(\sqrt[3]{x}\).