$\lim_{x\to 2} \, \sqrt{x-2}$ $$\lim_{x\to 2} \, \sqrt{x-2}$$
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.
 A: 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. 
