As you've probably noticed by now, both fractions' denominators turn out to be 0 when x = 1, so direct substitution is no good right now. However, we can combine the fractions first:
Notice now that we can safely cancel out the x - 1 term because, since this is a limit, we only care about what happens near 1 instead of at 1, and the fraction is defined near 1. Doing so gives the fraction [tex]\frac{1}{x + 1}[/tex], and substituting in 1 for x gives the answer [tex]\bf \frac{1}{2}[/tex].
