Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$
Solution 1
The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$. The reciprocal of $2$ is $1/2$, but $2$ is not less than $1/2$. The reciprocal of $-2$ is $-1/2$, and $-2$ is less than$-1/2$, so it is $\boxed{A}$.

Solution 2
The statement "a number is less than its reciprocal" can be translated as $x 0$, $x=0$, and $x0$, the sign of the inequality remains the same.  Thus, we have $x^2  0$.  This leads to $0  1$ when $x < 0$.  This leads to $x < -1$.
Putting the solutions together, we have $x<-1$ or $0 < x < 1$, or in interval notation, $(-\infty, -1) \cup(0, 1)$.  The only answer in that range is $\boxed {\text{(A)}\ -2}$

Solution 3
Starting again with $x < \frac{1}{x}$, we avoid multiplication by $x$.   Instead, move everything to the left, and find a common denominator:
$x < \frac{1}{x}$
$x - \frac{1}{x} < 0$
$\frac{x^2 - 1}{x} < 0$
$\frac{(x+1)(x-1)}{x} < 0$
Divide this expression at $x=-1$, $x=0$, and $x=1$, as those are the three points where the expression on the left will "change sign".
If $x<-1$, all three of those terms will be negative, and the inequality is true.  Therefore, $(-\infty, -1)$ is part of our solution set.
If $-1 < x < 0$, the $(x+1)$ term will become positive, but the other two terms remain negative.  Thus, there are no solutions in this region.
If $0 < x < 1$, then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative.  Thus, the entire region $(0, -1)$ is part of the solution set.
If $1 < x$, then all three terms are positive, and there are no solutions.
At all three "boundary points", the function is either $0$ or undefined.  Therefore, the entire solution set is $(-\infty, -1) \cup (0, 1)$, and the only option in that region is $x=-2$, leading to $\boxed{A}$.