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

It is given function is f(x) = (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, g(x) = and h(x) = x^ 2 First we have to prove that g(x) = and h(x) = 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 → k, then h → 0 ^ lim _ x g(x) = ^ lim _ x = ^ lim _ h -0 (k+h) = ^ lim _ h 0 -[ - . ] = ^ lim _ h -0 -^ lim _ h -0 = 0 - 0 = 1 - 0 = ^ lim _ x g(x) =g(k) Thus, g(x) = cosx is continuous function. Now, h(x) = x2 So, h is defined for every real number. Let c be a real number, then h(c) = c2 ^ lim _ x h(x) = ^ lim _ x x^ 2 ^ lim _ x h(x) =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(x2) is a continuous function.

Show that the function defined by f(x) = cos (x2 ) is a continuou…

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Question

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

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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 → k, then h → 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) = cosx is continuous function.
Now, h(x) = x²
So, h is defined for every real number.
Let c be a real number, then h(c) = c²
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{x}^{2}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} =\text{h(c)}$
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(x²) is a continuous function.