If $f(x)=\sqrt[\phantom{2}]{2+x^2},$ then the Linearization of $f(x)$ at $x=0$ is

Select one:

$L(x)=\sqrt[\phantom{2}]{2}$

$L(x)=\sqrt[\phantom{2}]{2}x$

$L(x)=x+\sqrt[\phantom{2}]{2}x$

$L(x)=\sqrt[\phantom{2}]{2}x+\sqrt[\phantom{2}]{2}$