Question

If \[f\left( x \right) = \left| {\begin{array}{*{20}{c}}
  {\cos x}&x&1 \\
  {2\sin x}&{{x^2}}&{2x} \\
  {\tan x}&x&1
\end{array}} \right|\] , then $\mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{x}$ is equal to:
A) Exists and is equal to $ - 2$
B) Does not exist
C) Exists and equal to $0$
D) Exists and is equal to $2$

Answer 1

Hint: The given function is in the form of determinant. We will first simplify the determinant and express the function in polynomial form. We will then see whether the function is differentiable or not. Then we will consider the given limit and simplify it by using properties of limits.

Complete step-by-step solution:
The given function is \[f\left( x \right) = \left| {\begin{array}{*{20}{c}}
  {\cos x}&x&1 \\
  {2\sin x}&{{x^2}}&{2x} \\
  {\tan x}&x&1
\end{array}} \right|\] .
The $3 \times 3$ determinant is solved as follows:
\[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a\left( {ei - fh} \right) - b\left( {di - gf} \right) + c\left( {dh - eg} \right)\]
Using this we can simplify the given function as follows:
$\Rightarrow$$f\left( x \right) = \cos x\left( {{x^2} - 2{x^2}} \right) - x\left( {2\sin x - 2x\tan x} \right) + 1\left( {2x\sin x - {x^2}\tan x} \right)$
Simplifying the above equation, we can write:
$\Rightarrow$$f\left( x \right) = {x^2}\left( {\tan x - \cos x} \right)$
Now observe that the obtained function is differentiable.
Differentiating with respect to $x$ we get,
$\Rightarrow$$f'\left( x \right) = 2x\left( {\tan x - \cos x} \right) + {x^2}\left( {{{\sec }^2}x + \sin x} \right)$
Now consider the given limit.
$\Rightarrow$$\mathop {\lim }\limits_{x \to 0} \dfrac{{2x\left( {\tan x - \cos x} \right) + {x^2}\left( {{{\sec }^2}x + \sin x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} 2\left( {\tan x - \cos x} \right) + x\left( {{{\sec }^2}x + \sin x} \right)$
Now clearly the function is a product of a polynomial and a trigonometric function so the limit clearly exists.
We will solve the limit using the properties of limits.
We will solve the above limit by direct substitution:
$\Rightarrow$$\mathop {\lim }\limits_{x \to 0} 2\left( {\tan x - \cos x} \right) + x\left( {{{\sec }^2}x + \sin x} \right) = 2\left( {\tan 0 - \cos 0} \right) + 0\left( {{{\sec }^2}0 + \sin 0} \right)$
Simplifying further we can write the following:
$\Rightarrow$$\mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{x} = - 2$
Therefore, the limit exists and is equal to $ - 2$ .

Thus, the correct option is A.

Note: The given function was expressed in terms of the determinant, we cannot directly differentiate the given function so we will simplify the determinant first then we will analyse the obtained function for differentiability. After that we solved the given limit. Before solving the limit, it is necessary to check whether the limit exists or not.