Show that the function defined by f(x)=cos(x2) is a continuous function.
The given function is f (x)=cos (x2)
This function f is defined for every real number and f can be written as the composition of two functions as,
f=goh, where g(x)=cosx and h(x)=x2[∵(goh)(x)=g(h(x))=g(x2)=cos(x2)=f(x)]
It has to be first proved that g(x)=cosx 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)=cosc
put x=c+h
If x\(\rightarrow\)c,then h\(\rightarrow\)0
\(\lim_{x\rightarrow c}\)g(x)=\(\lim_{x\rightarrow c}\) cosx
=\(\lim_{h\rightarrow 0}\) cos(c+h)
=\(\lim_{h\rightarrow 0}\)[cos c cos h-sin c sin h]
=\(\lim_{h\rightarrow 0}\) cos ccos 0-sin c sin 0
=cos c \(\times\)1-sin c \(\times\)0
=cos c
∴\(\lim_{x\rightarrow c}\)g(x)=g(c)
Therefore, g(x)=cos x is continuous function.
h(x)=x2
Clearly, h is defined for every real number.
Let k be a real number, then h(k)= k2
\(\lim_{x\rightarrow k}\)h(x)=\(\lim_{x\rightarrow k}\)x2=k2
∴\(\lim_{x\rightarrow k}\)h(x)=h(k)
Therefore, h is a continuous function.
It is known that for real-valued functions g and h, such that (goh) is defined at c,if g is continuous at c and if f is continuous at g(c),then (fog) is continuous at c.
Therefore,f(x)=(goh)(x)=cos(x2) is a continuous function.
The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 2+x, & \text{if } x \geq 0 \\ 2-x, & \text{if } x \leq 0 \end{cases} \] Then function \( f(x) \) at \( x=0 \) is: