Show that sec $x$ is a continuous function.

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Show that sec $x$ is a continuous function.

Solution:

Assume $f(x)=\sec x$
So, $f(x)=\frac{1}{\cos x}$
$f(x)$ is not defined when $\cos x=0$
And $\cos x=0$ when, $x=\frac{\pi}{2}$ and odd multiples of $\frac{\pi}{2}$ like $-\frac{\pi}{2}$
Let’s now consider the function
$f(a)=\cos a$ and let $c$ be any real number. Then,
$\begin{array}{l} \lim _{a \rightarrow c^{+}} f(a)=\lim _{h \rightarrow 0} f(c+h) \\ \lim _{h \rightarrow 0} \cos (c+h)=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\ =\cos c \lim _{h \rightarrow 0} \cosh -\sin c \lim _{h \rightarrow 0} \sin h \\ =\cos c(1)-\sin c(0) \end{array}$
So,
$\lim _{a \rightarrow c^{+}} f(a)=\cos c$
In the similar way,
$\lim _{a \rightarrow c^{-}} f(a)=f(c)=\cos c$
So,
$\lim _{a \rightarrow c^{-}} f(a)=\lim _{a \rightarrow c^{+}} f(a)=f(c)=\cos c$
Therefore, $\mathrm{f}(\mathrm{a})$ is continuous at $\mathrm{a}=\mathrm{c}$
In the similar way, $\cos \mathrm{x}$ is also continuous everywhere
Hence, $\sec x$ is continuous on the open interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$