(N/A) 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 $h(x)=x^{2}$.
$[\because (g \circ h)(x)=g(h(x))=g(x^{2})=\cos(x^{2})=f(x)]$.
It has to be first proved that $g(x)=\cos x$ and $h(x)=x^{2}$ 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 $x \to c$,then $h \to 0$.
$\lim_{x \to c} g(x) = \lim_{x \to c} \cos x = \lim_{h \to 0} \cos(c+h) = \lim_{h \to 0} [\cos c \cos h - \sin c \sin h]$.
$= \cos c \lim_{h \to 0} \cos h - \sin c \lim_{h \to 0} \sin h = \cos c \times 1 - \sin c \times 0 = \cos c$.
$\therefore \lim_{x \to c} g(x) = g(c)$. Therefore,$g(x)=\cos x$ is a continuous function.
Now,$h(x)=x^{2}$. Clearly,$h$ is defined for every real number. Let $k$ be a real number,then $h(k)=k^{2}$.
$\lim_{x \to k} h(x) = \lim_{x \to k} x^{2} = k^{2} = h(k)$. Therefore,$h$ is a continuous function.
It is known that if $g$ and $h$ are continuous functions,then their composition $(g \circ h)$ is also continuous.
Since $g(x)=\cos x$ and $h(x)=x^{2}$ are continuous,their composition $f(x)=(g \circ h)(x)=\cos(x^{2})$ is a continuous function.