-$\text{⨍(x)}$ is a continuous function satisfying

$\text{⨍ (x). ⨍ }\left(\frac{1}{\mathrm{x}}\right)\text{=⨍ (x)+⨍ }\left(\frac{1}{\mathrm{x}}\right)$ and $\text{⨍(1) > 0, }$ then

$\underset{\mathrm{x}\to 1}{\mathrm{Lt}}\text{⨍ (x)}$ is equal to

Concept: As it is given that f(x) is a continuous function.
This implies $\underset{x\to 1}{\lim }f\left(x\right)=f\left(1\right)$

We have given that $f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$
At $x=1$
$f\left(1\right)f\left(\frac{1}{1}\right)=f\left(1\right)+f\left(\frac{1}{1}\right)$

$\begin{array}{rcl}f\left(1\right)f\left(1\right)& =& f\left(1\right)+f\left(1\right)\\ f\left(1\right){]}^2& =& 2f\left(1\right)\\ f\left(1\right)\left(f\left(1\right)-2\right)& =& 0\\ & & \\ f\left(1\right)f\left(1\right)& =& 2f\left(1\right)\\ & & \\ f\left(1\right)& =& 2 f\left(1\right)=0\Rightarrow 0=0\\ \text{ Thus, }\underset{x\to 1}{\lim }f\left(x\right)& =& f\left(1\right)=2\end{array}$

Given $f\left(1\right)>0$, so that $f\left(1\right)=2$

Hence, the correct answer is 2.