If $f(x)=\left|\begin{array}{ccc}\sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2 x & 1 & 1\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}$ is

(A) 3

(B) -1

(C) 0

(D) 1

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Best answer

**SOLUTION —**

$f(x)=x(x-1) \sin x-\left(x^3-2 x^2\right) \cos x-x^3 \tan x$

$=x^2 \sin x-x^3 \cos x-x^3 \tan x+2 x^2 \cos x-x \sin x$ $\therefore \lim _{x \rightarrow 0} \frac{f(x)}{x^2}$

$=\lim _{x \rightarrow 0}\left(\sin x-x \cos x-x \tan x+2 \cos x-\frac{\sin x}{x}\right)$ $=0-0-0+2-1=1$

So, The correct option will be **(D).**

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