Question

If the function $f\left(x\right)=\frac{\left[\cos \left(\sin x\right)-\cos x\right]}{x^4}$ is continuous at each point in its domain and $f\left(0\right)=\frac{1}{k}$, then $k$ is:

Answer 1

Step-1 Applying the condition for continuity:

It is given that the function is continuous at each point in its domain, so $\underset{x\to 0}{\lim }\frac{\left[\cos \left(\sin x\right)-\cos x\right]}{x^4}=f\left(0\right)$

$\begin{array}{ccc}\Rightarrow & \underset{x\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{\sin x+x}{2}}\right)\sin \left({\displaystyle \frac{\sin x-x}{2}}\right)}{x^4}=\frac{1}{k}& [\because \mathrm{cosa}-\mathrm{cosb}=-2\sin \left(\frac{a+b}{2}\right)\sin \left(\frac{a-b}{2}\right)]\\ \Rightarrow & \underset{x\to 0}{\lim }\frac{2\sin \left({\displaystyle \frac{\sin x+x}{2}}\right)\sin \left({\displaystyle \frac{\sin x-x}{2}}\right)}{x^4}·\frac{\left({\displaystyle \frac{\sin x+x}{2}}\right)}{\left({\displaystyle \frac{\sin x+x}{2}}\right)}·\frac{\left({\displaystyle \frac{\sin x-x}{2}}\right)}{\left({\displaystyle \frac{\sin x-x}{2}}\right)}=\frac{1}{k}& \\ & & \end{array}$

Step-2 Solving limit for the value of $k$:

Let, $\alpha =\frac{\sin x+x}{2}$ and $\beta =\frac{\sin x-x}{2}$

$\begin{array}{ccc}\Rightarrow & \underset{x\to 0}{\lim }\frac{-2\sin \left(\alpha \right)\sin \left(\beta \right)}{x^4}·\frac{\alpha }{\alpha }·\frac{\beta }{\beta }=\frac{1}{k}& \\ \Rightarrow & -\underset{x\to 0}{\lim }\left(\frac{2}{x^4}\right)·\frac{\sin \left(\alpha \right)}{\alpha }·\frac{\sin \left(\beta \right)}{\beta }·\frac{\alpha \beta }{1}=\frac{1}{k}& \\ \Rightarrow & -\underset{x\to 0}{\lim }\frac{2}{x^4}·\alpha \beta =\frac{1}{k}& [\because \underset{\theta \to 0}{lim}\frac{sin\theta }{\theta }=1]\\ \Rightarrow & -\underset{x\to 0}{\lim }\left(\frac{2}{x^4}\right)·\left(\frac{\sin x+x}{2}\right)\left(\frac{\sin x-x}{2}\right)=\frac{1}{k}& [\because \alpha =\frac{\mathrm{sinx}+x}{2},\beta =\frac{\mathrm{sinx}-x}{2}]\\ \Rightarrow & -\underset{x\to 0}{\lim }\left(\cancel{2}\right)\left(\frac{\sin x+x}{\cancel{2}x}\right)\left(\frac{\sin x-x}{2x^3}\right)=\frac{1}{k}& \\ \Rightarrow & -\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}+\frac{\cancel{x}}{\cancel{x}}\right)\left(\frac{1}{2}\right)\left(\frac{\sin x-x}{2x^3}\right)=\frac{1}{k}& \\ \Rightarrow & -\underset{x\to 0}{\lim }\left(1+1\right)\left(\frac{1}{2}\right)\left(\frac{\sin x-x}{x^3}\right)=\frac{1}{k}& \end{array}$

Step-3 Using Maclaurin series for the limit:

We know that the Maclaurin series expansion of $\sin x$ is
$\begin{array}{cc}& \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots \\ \Rightarrow & \frac{\sin x-x}{x^3}=\frac{\left(x-{\displaystyle \frac{x^3}{3!}}+{\displaystyle \frac{x^5}{5!}}-{\displaystyle \frac{x^7}{7!}}+\dots \right)-x}{x^3}\\ \Rightarrow & \frac{\sin x-x}{x^3}=-\frac{1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\dots \end{array}$

Thus,

$\begin{array}{cc}\Rightarrow & -\underset{x\to 0}{\lim }\left(\cancel{1+1}\right)\left(\frac{1}{\cancel{2}}\right)\left(-\frac{1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\dots \right)=\frac{1}{k}\\ \Rightarrow & -\left(-\frac{1}{3!}+\frac{0^2}{5!}-\frac{0^4}{7!}+\dots \right)=\frac{1}{k}\\ \Rightarrow & \frac{1}{6}=\frac{1}{k}\\ \therefore & k=6\end{array}$

Therefore, the value of $k$ is $6$.