Question

Let $f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x& 2x\\ \sin x& x& x\end{array}\right|$, then $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^2}$ is equal to

(a) 0
(b) −1
(c) 2
(d) 3

Answer 1

$$\begin{gathered} f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x& 2x\\ \sin x& x& x\end{array}\right| \\ =\left|\begin{array}{ccc}\cos x& x& 1\\ \sin x& 0& x\\ \sin x& x& x\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_3\right] \\ =\left|\begin{array}{ccc}\cos x& x& 1\\ \sin x& 0& x\\ \sin x-\cos x& 0& x-1\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-R_1\right] \\ =-x\left[x\sin x-\sin x-x\sin x+x\cos x\right] \\ =-x\left(x\cos x-\sin x\right) \\ \therefore \underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^2}=\underset{x\to 0}{\lim }\frac{x\left(\sin x-x\cos x\right)}{x^2} \\ =\underset{x\to 0}{\lim }\frac{\sin x}{x}-\underset{x\to 0}{\lim }\cos x \\ =1-1=0 \end{gathered}$$

Hence, the correct option is (a).