Because the expression is not real-valued left of #2#, a beginning calculus course will say there is no (two-sided) limit as #x# approaches #2#. We can find the limit as #x# approaches #2# from the right. (Thanks George C., for pointing this out.)
#(sqrtx-sqrt2)/sqrt(x^2-2x)# as #xrarr2^+#, this expression has indeterminate form #0/0#.
#x^2-2x# is not a perfect square. I could complete the square to get #x^2-2x+1-1 = (x-1)^2-1#. That's not a perfect square. It is a difference of squares, which might be helpful for some problems, but I don't see a way to use it here.
We could write #sqrt(x^2-2x) = sqrt(x^2)sqrt(1-2/x)# #" "# (for #x != 0#). That doesn't look promising.
We can rationalize the numerator by multiplying by #(sqrtx+sqrt2)/(sqrtx+sqrt2)#.
We get
#(x-2)/(sqrt(x^2-2x)(sqrtx+sqrt2))# The form of the limit as #xrarr2^+# is still indeterminate, but wait there's more I could do now. I have something involving #x-2# in the denominator.
Factor #sqrtx# out of #sqrt(x^2-2x)#
#(x-2)/(sqrt(x^2-2x)(sqrtx+sqrt2)) = (x-2)/(sqrtxsqrt(x-2)(sqrtx+sqrt2))#
Now reduce the fraction using #a/sqrta = (sqrta)^2/sqrta = (cancel(sqrta)sqrta)/cancel(sqrta) = sqrta#
#(x-2)/(sqrtxsqrt(x-2)(sqrtx+sqrt2)) = sqrt(x-2)/(sqrtx(sqrtx+sqrt2))#.
The limit as #xrarr2^+# is #0/(sqrt2(sqrt2+sqrt2)) = 0#.
Presenting it without the false starts and explanation looks like:
#lim_(xrarr2^+)(sqrtx-sqrt2)/sqrt(x^2-2x) = lim_(xrarr2^+)(x-2)/(sqrtxsqrt(x-2)(sqrtx+sqrt2))#
# = lim_(xrarr2^+)sqrt(x-2)/(sqrtx(sqrtx+sqrt2))#
# = 0/(sqrt2(sqrt2+sqrt2)) = 0#
