Official solution
Given,
\[\mathop {\lim }\limits_{x \to 3} \] \[{{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}\]
Here if you put x = 3 in \[{{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}\]
you will get \[{0 \over 0}\] form.
So, we can apply L' Hospital rule
\[\therefore\,\,\,\] \[\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}\]
= \[\mathop {\lim }\limits_{x \to 3} \] \[{{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}\] (applying L' Hospital rule)
= \[{{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}\]
= \[{1 \over 2}\] \[ \times \] \[\sqrt 2 \]
= \[{1 \over {\sqrt 2 }}\]