What is the limit of a complex fraction with L'Hopital's rule?

[karush]

\tiny{s8.1.6.64}
Evaluate
$\displaystyle\lim_{x \to 2}
\dfrac{\sqrt{6-x}{-2}}{\sqrt{3-x}-1}$

ok so if you plug in 2 directly you get $\dfrac{0}{0}$

So we either use L'H rule or use conjugate

or is there better way

 

[skeeter]

L'Hopital works

you can also multiply numerator and denominator by both conjugates