$$\lim_{x\to 2} \, \sqrt{x-2}$$
When you take the right hand limit for this expression, you get $0$. However, if you take the left hand side it gives an imaginary number.
However, do you consider the imaginary part when taking the limit (in which case both sides would tend to $0$) or do you consider the limit to be undefined because it cannot be approached from the left in the field of real numbers.
Well, this is tricky. Part of the issue is what does the symbol $\sqrt{\cdot}$ mean? If $a$ is a positive real number, we denote by $\sqrt{a}$ the positive real number whose square is $a$. Usually, if we want a square root of a negative or complex number, we have to be very specific about which square root. For example, we usually say $i$ is the square root of $-1$, but this isn't quite right, since $-i$ is also a square root of $-1$.
In calculus, when dealing with functions of real variables, we don't want to get into this sticky territory, so we say that the domain of the function $f(x) = \sqrt{x}$ is the set of non-negative reals. Therefore, if we want to take a limit as $x$ goes to $0$, we can only consider paths to $0$ that lie within the domain of the function. In this case, that means we can only take a 'right hand' limit.
Interestingly, though, we could think of $f$ as a complex valued function, so taking a 'left hand' limit would give \begin{eqnarray*} \lim_{x\to 2^{-}} \sqrt{x-2} &=& \lim_{x\to 2^{-}} \\ &=&\lim_{x\to 2^-} \sqrt{(-1)(2-x)}\\ &=& \lim_{x\to 2^{-}} i\;\sqrt{2-x}\\ &=& i \cdot \lim_{x\to 2^{-}} \sqrt{2-x}\\ &=& 0, \end{eqnarray*} so, in a certain sense, everything checks out.