Show that the function defined by f (x) = cos (x2) is a continuous function.

Solution:

The given function is $f(x)=\cos \left(x^{2}\right)$

This function f is defined for every real number and f can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=\cos x$ and $h(x)=x^{2}$

$\left[\because(g o h)(x)=g(h(x))=g\left(x^{2}\right)=\cos \left(x^{2}\right)=f(x)\right]$

It has to be first proved that g (x) = cos x and h (x) = x2 are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g (c) = cos c

Put $x=c+h$

If $x \rightarrow c$, then $h \rightarrow 0$

$\begin{aligned} \lim _{x \rightarrow c} g(x) &=\lim _{x \rightarrow c} \cos x \\ &=\lim _{h \rightarrow 0} \cos (c+h) \\ &=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\ &=\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\ &=\cos c \cos 0-\sin c \sin 0 \\ &=\cos c \times 1-\sin c \times 0 \\ &=\cos c \end{aligned}$

$\therefore \lim _{x \rightarrow c} g(x)=g(c)$

Therefore, g (x) = cos x is continuous function.

$h(x)=x^{2}$

Clearly, h is defined for every real number.

Let $k$ be a real number, then $h(k)=k^{2}$

$\lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k} x^{2}=k^{2}$

$\therefore \lim _{x \rightarrow k} h(x)=h(k)$

Therefore, h is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, $f(x)=(g o h)(x)=\cos \left(x^{2}\right)$ is a continuous function.