#lim(sqrt(3+x)-(sqrt(3))/x),x,0#
You can think of this limit in two parts: first, as a limit goes to zero of #sqrt(3+x)# and second as a limit goes to zero of #(sqrt(3))/x#.
It is clear that limit for the first part would be just #sqrt(3)#. Since, the #sqrt(3)# won't have any significant contribution to the limit, we can easily modify the question from #lim(sqrt(3+x)-(sqrt(3))/x)# to #lim(-sqrt(3)/x)#.
Now, we can find the limit of the function #-sqrt(3)/x# as x approaches zero from negative side and we note that the smaller and smaller negative number that is very close to zero but it's not zero would produce really big numbers. So, the limit as x goes to zero from negative direction is positive infinity. Similarly, the limit as x goes to zero from positive direction is negative infinity.
Since the limits from left and right directions are not the same, the limit of this function is said to be does not exist.
