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 $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}$ is

• A. $-1$
• B. $3$
• C. $1$
• D. Zero

Answer 1

Answer: (C)

$1$

$f\left(x\right)=\left|\begin{array}{ccc}sinx& cosx& tanx\\ x^3& x^2& x\\ 2x& 1& 1\end{array}\right|⟹f\left(x\right)=sinx(x^2-x)-cosx(x^3-2x^2)+tanx(x^3-2x^3)⟹f\left(x\right)=x^{2}sinx-xsinx-x^{3}cosx+2x^{2}cosx-x^{3}tanx⟹\frac{f\left(x\right)}{x^2}=sinx-xcosx+2cosx-xtanx-\frac{sinx}{x}⟹\underset{x⟶0}{lim}\frac{f\left(x\right)}{x^2}=\underset{x⟶0}{lim}sinx-\underset{x⟶0}{lim}xcosx+\underset{x⟶0}{lim}2cosx-\underset{x⟶0}{lim}xtanx-\underset{x⟶0}{lim}\frac{sinx}{x}⟹\underset{x⟶0}{lim}\frac{f\left(x\right)}{x^2}=0-0+2-0-1⟹\underset{x⟶0}{lim}\frac{f\left(x\right)}{x^2}=1$