Evaluating $h\left(x\right)$ at $x\=0$ will clearly result in a "division by zero" error. Note also that the numerator also evaluates to zero at $x\=0$. Hence, the function $h$ is not defined at $x\=0$. But even though it is a fraction, $h$ is not a rational function because it is not a ratio of two polynomials. Thus, the Factor and Remainder theorem does not apply. Yet, there may still be an algebraic transformation that results in the cancellation of a common factor of $x-0$. Indeed, there is, and this is the device of rationalizing the numerator.

To rationalize the numerator, multiply the fraction by 1 in the form $\frac{\sqrt{1\+x}\+1}{\sqrt{1\+x}\+1}$, where $\sqrt{1\+x}\+1$ is referred to as the conjugate of $\sqrt{1\+x}-1$.

Because all five expressions on the right in Table 1.1.4(a) are equal to $h\left(x\right)$, the restriction that $x\ne 0$ applies to all. However, the last one, regarded in its own right, can indeed be evaluated at $x\=0$, and the resulting value, namely, $1\/2$, is the required limit. In other words, by transforming $h\left(x\right)$ by rationalizing its numerator, an expression results that extends the domain of $h$. The value of the extended function at $x\=0$ tells the behavior of $h\left(x\right)$ near $x\=0$.

Unfortunately, if $h\left(x\right)$ is multiplied by this form of 1, Maple would reduce the multiplier to 1 immediately, and $h\left(x\right)$ would be multiplied by just 1. In Maple, the numerator and denominator of $h\left(x\right)$ must be separately multiplied by the "conjugate" of its numerator, as per the following calculation.