Question

If the function $f(x) = \dfrac{{\cos (\sin x) - \cos x}}{{{x^4}}}$ is continuous at each point in its domain and $f(0) = \dfrac{1}{k}$, then what is the value of $k$ ?

Answer 1

Hint: First, the function $f(x)$ is simplified by using the trigonometric identities and then the numerical value of $\mathop {\lim }\limits_{x \to 0} f(x)$ is calculated by using the limit formulae, L’ Hospital Rule and equated to given limit $\dfrac{1}{k}$ to find the value of $k$.

Formula Used:
In the first step, the trigonometric identity used to simplify the function is
$\cos B - \cos A = 2\sin \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}$ .
In the second step, the limit formulae used is $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ .
In the third step, L’ Hospital Rule is used, according to which, if a function $f(x)$ is in the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ at a point, then its numerator and denominator are differentiated with respect to $x$ and then the limit at that point is calculated.

Complete step by step solution:
We have been given that the function $f(x) = \dfrac{{\cos (\sin x) - \cos x}}{{{x^4}}}$ is continuous at each point in its domain and $f(0) = \dfrac{1}{k}$ .
We know that a function at a point is continuous, if the following three situations are satisfied:-
First, the limit must exist at that point.
Second, the function must be defined at that point.
Third, the limit and function must have equal values at that point.
Here, the function $f(x)$ is already defined and continuous and $f(0) = \dfrac{1}{k}$ as per given data.
So, we will examine its limit as .
First, we will simplify the function $f(x) = \dfrac{{\cos (\sin x) - \cos x}}{{{x^4}}}$.
As per trigonometric identity $\cos B - \cos A = 2\sin \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}$.
Let, $x = A$ and $\sin x = B$
Then,
$\cos (\sin x) - \cos x = \cos B - \cos A$
$ = 2\sin \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}$ [Using the trigonometric identity]
$ = 2\sin \dfrac{{x + \sin x}}{2}\sin \dfrac{{x - \sin x}}{2}$ (Putting the values of A and B]
Now, we will find the limit of the function.
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos (\sin x) - \cos x}}{{{x^4}}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin \dfrac{{x + \sin x}}{2}\sin \dfrac{{x - \sin x}}{2}}}{{{x^4}}}$ [Putting the simplified value of $\{ \cos (\sin x) - \cos x\} $]
Multiplying $\dfrac{{x + \sin x}}{2}$ and $\dfrac{{x - \sin x}}{2}$ in both numerator and denominator, we have

$\mathop {\lim }\limits_{x \to 0} \dfrac{2}{{{x^4}}} \times \dfrac{{\sin \dfrac{{x + \sin x}}{2}}}{{\dfrac{{x + \sin x}}{2}}} \times \dfrac{{\sin \dfrac{{x - \sin x}}{2}}}{{\dfrac{{x - \sin x}}{2}}} \times \dfrac{{x + \sin x}}{2} \times \dfrac{{x - \sin x}}{2}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - {{\sin }^2}x}}{{2{x^4}}} \times \dfrac{{\sin \dfrac{{x + \sin x}}{2}}}{{\dfrac{{x + \sin x}}{2}}} \times \dfrac{{\sin \dfrac{{x - \sin x}}{2}}}{{\dfrac{{x - \sin x}}{2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - {{\sin }^2}x}}{{2{x^4}}} \times \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{{x + \sin x}}{2}}}{{\dfrac{{x + \sin x}}{2}}} \times \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{{x - \sin x}}{2}}}{{\dfrac{{x - \sin x}}{2}}}$ [Since $\mathop {\lim }\limits_{x \to 0} f(x) \times g(x) = \mathop {\lim }\limits_{x \to 0} f(x) \times \mathop {\lim }\limits_{x \to 0} g(x)$]
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - {{\sin }^2}x}}{{2{x^4}}} \times 1 \times 1$ [Using $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$]

Further, simplifying it, we have
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - {{\sin }^2}x}}{{2{x^4}}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2x - 2\sin x\cos x}}{{8{x^3}}}$ [Using L’ Hospital Rule numerator and denominator differentiated due to $\dfrac{0}{0}$ form]
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2x - \sin 2x}}{{8{x^3}}}$ [Since $2\sin x\cos x = \sin 2x$]
Again using L’ Hospital Rule due to $\dfrac{0}{0}$ form, we have
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{2 - 2\cos 2x}}{{24{x^2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4\sin 2x}}{{48x}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{8\cos 2x}}{{48}}$ [Using L’ Hospital Rule]
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 2x}}{6}$
Simplifying further
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 2x}}{6} = \dfrac{{\cos 0}}{6} = \dfrac{1}{6}$
As $f(x)$ is a continuous function at each point of its domain, $\mathop {\lim }\limits_{x \to 0} f(x) = f(0) = \dfrac{1}{6}$
But, it is given that $f(0) = \dfrac{1}{k}$
So, $\begin{array}{l}\dfrac{1}{k} = \dfrac{1}{6}\\ \Rightarrow k = 6\end{array}$
Thus, the value of $k$ is equal to $6$.

Hence, the value of $k$ is equal to $6$.

Note: A function must be defined at a point, its limit must exist at that place, and the value of the function at that position must equal the value of the limit at that point for it to be continuous at that point. If these conditions are not met, discontinuity can seem as detachable, leap, or infinite.