Question
Show that the function defined by $f(x) = \cos (x^2 )$ is a continuous function.
✓
Answer
It is given function is $\text{f(x)} = \cos(\text{x}^{2})$
This function f is defined for every real number and $f$ can be written as the composition of two function as,
$f =$ goh, where, $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$
First we have to prove that $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$ are continuous functions.
We know that g is defined for every real number.
Let $k$ be a real number.
Then, $g(k) = \cos k$
Now, put $x = k + h$
lf $x \rightarrow k,$ then $h \rightarrow 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\cos\text{x}$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos (\text{k}+\text{h})$
$= ^{\text{lim}}_{\text{h}\rightarrow\text{0}}-[\cos\text{k}\cos\text{h} - \sin\text{k}.\sin\text{h}]$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos\text{k}\cos\text{h} -^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}} \sin\text{k}\sin\text{h}$
$= \cos\text{k}\cos0 - \sin\text{k}\sin0$
$= \cos\text{k} \times 1 - \sin \times\ 0$
$= \cos\text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
Thus, $g(x) = \cos x$ is continuous function.
Now, $h(x) = x^2$
So, $h$ is defined for every real number.
Let $c$ be a real number, then $h(c) = c^2$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{x}^{2}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} =\text{h(c)}$
Therefore, $h$ is a continuous function.
We know that for real valued functions $g$ and $h,$ Such that $($fog$)$ is continuous at $c.$
Therefore, $f(x) = (goh)(x) = \cos(x^2)$ is a continuous function.
This function f is defined for every real number and $f$ can be written as the composition of two function as,
$f =$ goh, where, $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$
First we have to prove that $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$ are continuous functions.
We know that g is defined for every real number.
Let $k$ be a real number.
Then, $g(k) = \cos k$
Now, put $x = k + h$
lf $x \rightarrow k,$ then $h \rightarrow 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\cos\text{x}$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos (\text{k}+\text{h})$
$= ^{\text{lim}}_{\text{h}\rightarrow\text{0}}-[\cos\text{k}\cos\text{h} - \sin\text{k}.\sin\text{h}]$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos\text{k}\cos\text{h} -^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}} \sin\text{k}\sin\text{h}$
$= \cos\text{k}\cos0 - \sin\text{k}\sin0$
$= \cos\text{k} \times 1 - \sin \times\ 0$
$= \cos\text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
Thus, $g(x) = \cos x$ is continuous function.
Now, $h(x) = x^2$
So, $h$ is defined for every real number.
Let $c$ be a real number, then $h(c) = c^2$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{x}^{2}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} =\text{h(c)}$
Therefore, $h$ is a continuous function.
We know that for real valued functions $g$ and $h,$ Such that $($fog$)$ is continuous at $c.$
Therefore, $f(x) = (goh)(x) = \cos(x^2)$ is a continuous function.
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