Let $f:\left( {0,\infty } \right) \to R$ be a differentiable function such that $f'\left( x \right) = 2 - \dfrac{{f\left( x \right)}}{x}$ for all $x$ and $x \in \left( {0,\infty } \right)$ and $f\left( 1 \right) \ne 1$. Then:
A) \[\mathop {\lim }\limits_{x \to {0^ + }} f\left( {\dfrac{1}{x}} \right) = 1\]
B) \[\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {\dfrac{1}{x}} \right) = 2\]
C) \[\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'\left( x \right) = 0\]
D) \[\mathop {\lim }\limits_{x \to 0} f'\left( {\dfrac{1}{x}} \right) = 1\]

Hint:Rewrite the equation in general form and compare with the first order differential equation and find the integrating factor and substitute the value in general equation of differential equation. Apply the limit and find the value.

Complete step by step solution:
According to the question it is given that the equation of differentiable is $f'\left( x \right) = 2 - \dfrac{{f\left( x \right)}}{x}$ for all $x$ and the condition is $f\left( 1 \right) \ne 1$.
The differential equation is the topic of calculus in mathematics. It is used to find the exact value of any situation within the specified limit. There are various types of topics in differential equations. First order differential equation is one of them.
Let $\dfrac{{dy}}{{dx}} + Px = Q$ be a first order differential equation. Then the integrating factor for the equation can be found by using the formula which is written below,
$I \cdot F = {e^{\int {P\left( x \right)dx} }}$
The solution of the equation can be written as $f\left( x \right) \cdot \left( {I \cdot F} \right) = \int {Q\left( x \right) \cdot I \cdot Fdx} $. The resultant equation is the solution of the differential equation.
Given,
Consider the equation.
$f'\left( x \right) = 2 - \dfrac{{f\left( x \right)}}{x}$
Rewrite the above equation.
$f'\left( x \right) + \dfrac{{f\left( x \right)}}{x} = 2$
Compare the above equation with the general equation $\dfrac{{dy}}{{dx}} + Px = Q$.
So, $P\left( x \right) = \dfrac{1}{x}$ and $Q = 2$.
Substitute $P\left( x \right) = \dfrac{1}{x}$ in the integrating factor $I \cdot F = {e^{\int {P\left( x \right)dx} }}$.
$
  I \cdot F = {e^{\int {\dfrac{1}{x}dx} }} \\
   = {e^{\ln x}} \\
   = x \\
 $
Substitute $x$ for $I.F$ in the equation $f\left( x \right) \cdot \left( {I \cdot F} \right) = \int {Q\left( x \right) \cdot I \cdot Fdx} $.
$
  f\left( x \right) \cdot \left( {I \cdot F} \right) = \int {Q\left( x \right) \cdot I \cdot Fdx} \\
  xf\left( x \right) = \int {2xdx} \\
  xf\left( x \right) = {x^2} + C \\
 $
As $f\left( 1 \right) \ne 1$, substitute 1 for $x$ in the equation $xf\left( x \right) = {x^2} + C$ to find the value of $C$.
$
  \left( 1 \right)\left( 1 \right) \ne {\left( 1 \right)^2} + C \\
  C \ne 0 \\
 $
Differentiate the function $xf\left( x \right) = {x^2} + c$ with respect to $x$.
\[
  \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left[ {x + \dfrac{c}{x}} \right] \\
  f'\left( x \right) = 1 - \dfrac{c}{{{x^2}}} \\
 \]
Apply the limit as $x$ approaches to 0.
\[
  \mathop {\lim }\limits_{x \to 0} f'\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {1 - \dfrac{c}{{{x^2}}}} \right) \\
   = 1 \\
 \]
It implies that,
\[
  \mathop {\lim }\limits_{x \to 0} f'\left( {\dfrac{1}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {1 - c{x^2}} \right) \\
   = 1 \\
 \]

So, from the above calculation it is concluded that the value of \[\mathop {\lim }\limits_{x \to 0} f'\left( {\dfrac{1}{x}} \right) = 1\].

Hence, the option (A) is correct.

Note:Make sure that the applied limit is correct and there should not be any calculation mistake. Make sure that the integrating factor contains the $P\left( x \right)$ term, not the $Q$.