Question
Show that $\text{f(x)}=\cos\text{x}^2$ is a continuous function.
✓
Answer
Given, $\text{f(x)}=\cos\big(\text{x}^2\big)$
This function f is defined for every real number and f can be written as the composition of two functions as
f = goh, where $\text{g(x)}=\cos\text{x}$ and h(x) =$ x^2$
$\big[\because(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\text{x}^2)=\cos(\text{x}^2)=\text{f(x)}\big]$
It has to be first proved that $\text{g(x)}=\cos\text{x}$ 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, $\text{g(c)}=\cos\text{c}$
Put x = c + h
If x → c, then h→ 0
$=\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x})=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
So, $\text{g(x)}=\cos\text{x}$ 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$
$=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{x}^2=\text{k}^2$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\text{h}(\text{k})$
So, h is a continuous function.
It is known that for real valued functions g and h, such that (goh) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (fog) is continuous at x = c.
Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\cos(\text{x}^2)$ is a continuous function.
This function f is defined for every real number and f can be written as the composition of two functions as
f = goh, where $\text{g(x)}=\cos\text{x}$ and h(x) =$ x^2$
$\big[\because(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\text{x}^2)=\cos(\text{x}^2)=\text{f(x)}\big]$
It has to be first proved that $\text{g(x)}=\cos\text{x}$ 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, $\text{g(c)}=\cos\text{c}$
Put x = c + h
If x → c, then h→ 0
$=\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x})=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
So, $\text{g(x)}=\cos\text{x}$ 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$
$=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{x}^2=\text{k}^2$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\text{h}(\text{k})$
So, h is a continuous function.
It is known that for real valued functions g and h, such that (goh) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (fog) is continuous at x = c.
Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\cos(\text{x}^2)$ is a continuous function.
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