Question

$\Delta \left(x\right)=\left|\begin{array}{ccc}\tan x& \tan (x+h)& \tan (x+2h)\\ \tan (x+2h)& \tan x& \tan (x+h)\\ \tan (x+h)& \tan (x+2h)& \tan x\end{array}\right|$
Find the value of $\underset{h\to 0}{lim}\frac{\sqrt{3}\Delta \left(\pi /3\right)}{h^2}$

Answer 1

Using $C_3\to C_3-C_2$ and $C_2\to C_2-C_1$, we can write
$\frac{\Delta }{h^2}=\left|\begin{array}{ccc}\tan x& \left[\tan \right(x+h)-\tan x]/h& \left[\tan \right(x+2h)-\tan (x+h\left)\right]/h\\ \tan (x+2h)& [\tan x-\tan (x+2h\left)\right]/h& \left[\tan \right(x+h)-\tan x]/h\\ \tan (x+h)& \left[\tan \right(x+2h)-\tan (x+h\left)\right]/h& [\tan x-\tan (x+2h\left)\right]/h\end{array}\right|$
But
$\underset{h\to 0}{lim}\frac{\tan (x+h)-\tan x}{h}$
$=\underset{h\to 0}{lim}\frac{1}{h}\frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h)\cos x}$
$=\underset{h\to 0}{lim}\frac{1}{h}\frac{\sin h}{\cos (x+h)\cos x}={\sec }^{2}x$ or (use $\frac{d}{dx}\left(\tan x\right)={\sec }^{2}x$)
Similarly, $=\underset{h\to 0}{lim}\frac{\tan (x+2h)-\tan (x+h)}{h}={\sec }^{2}x$
and $=\underset{h\to 0}{lim}\frac{\tan x-\tan (x+2h)}{h}=-2{\sec }^{2}x$
Thus,
$\frac{\Delta \left(x\right)}{h^2}=\left|\begin{array}{ccc}\tan x& {\sec }^{2}x& {\sec }^{2}x\\ \tan x& -{\sec }^{2}x& {\sec }^{2}x\\ \tan x& {\sec }^{2}x& -2{\sec }^{2}x\end{array}\right|$ $=\left|\begin{array}{ccc}\tan x& {\sec }^{2}x& {\sec }^{2}x\\ 0& -3{\sec }^{2}x& 0\\ 0& 0& -3{\sec }^{2}x\end{array}\right|$
$=9\tan x{\sec }^{4}x$ [Using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$]
So $\underset{h\to 0}{lim}\frac{\sqrt{3}\Delta \left(\pi /3\right)}{h^2}=9\times 3\times 2^4=216$