If f(x)= | … |, then lim _x → 0 f^prime(x) x

$$\begin{aligned} & f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right| \\=& \cos x\left(x^2-2 x^2\right)-x(2 \sin x-2 x \tan x) \\=&-x^2 \cos x-2 x \sin x+2 x^2 \tan x \\\left.+2 x \sin x-x^2 \tan x\right) \\=& x^2 \tan x-x^2 \tan x \\ \Rightarrow & f^{\prime}(x)=2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right) \\ \therefore & \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x} \\=& \lim _{x \rightarrow 0} \frac{2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right)}{x} \\=& \lim _{x \rightarrow 0}(\tan x-\cos x)+x\left(\sec ^2 x+\sin x\right) \\=& 2(0-1)+0=-2 \\ \text { So, } \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}=-2 \end{aligned}$$==+2xsinx−x2tanx)=⇒∴=== So, x→0lim​xf′(x)​=−2​f(x)=​cosx2sinxtanx​xx2x​12x1​​cosx(x2−2x2)−x(2sinx−2xtanx)−x2cosx−2xsinx+2x2tanxx2tanx−x2tanxf′(x)=2x(tanx−cosx)+x2(sec2x+sinx)x→0lim​xf′(x)​x→0lim​x2x(tanx−cosx)+x2(sec2x+sinx)​x→0lim​(tanx−cosx)+x(sec2x+sinx)2(0−1)+0=−2​