Question

If $f\left(x\right)=\left|\begin{array}{ccc}\sin x& \cos x& \tan x\\ x^3& x^2& x\\ 2x& 1& 1\end{array}\right|,$ then the value of $\underset{x\to 0}{lim}\left(\frac{f\left(x\right)}{x^2}\right)$

Answer 1

$f\left(x\right)=\left|\begin{array}{ccc}\sin x& \cos x& \tan x\\ x^3& x^2& x\\ 2x& 1& 1\end{array}\right|\text{Expanding the determinant along the first row,}\Rightarrow f\left(x\right)=\sin x(x^2-x)-\cos x(x^3-2x^2)+\tan x(x^3-2x^3)\therefore \underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}=\frac{\sin x(x^2-x)-\cos x(x^3-2x^2)-x^3\tan x}{x^2}=\underset{x\to 0}{lim}(\sin x-\frac{\sin x}{x}-x\cos x+2\cos x-x\tan x)=1$