We have two square root functions. The restriction that exists for the typical square root function #f(x)=sqrtx# is that #x>=0#.
So, we know that both #x-2>=0# and #x^2-4x-4>=0# for this function to be defined.
The first can be easily manipulated to show that #x>=2#. Since we want the limit as #x# approaches #x=2#, this isn't a problem yet.
To solve the quadratic inequality, first consider #x^2-4x-4=0#. This is solvable through the quadratic formula or completing the square to show that this quadratic has no real solutions, thus it never crosses the #x#-axis. Testing any #x#-value shows that this function will always lie under the #x#-axis and never is greater than or equal to #0#.
Thus, due to the function #sqrt(x-2)/sqrt(x^2-4x-4)# not existing at #x=2#, or any other point, the limit does not exist.
Please comment if you meant to say #sqrt(x-2)/sqrt(x^2-4x+4)# instead.
