Consider lim_{x→0^{+}}(x ln x). This is an indeterminate form of the type 0·∞. To apply l’Hôpital’s rule we must rewrite it as a quotient.

First try: lim_{x→0^{+}} \frac{x}{(ln x)^{−1}} is an indeterminate form of type \frac{0}{0} . The rule gives

lim_{x→0^{+}} \frac{1}{−\frac{1/x}{(ln x)^{2}}} = lim_{x→0^{+}}[−x(ln x)^{2}],

which is more complicated than the original problem.

Second try: lim_{x→0^{+}} \frac{ln x}{1/x} is an indeterminate form of type \frac{∞}{∞} . The rule gives

lim_{x→0^{+}} \frac{1/x}{−1/x^{2}} = lim_{x→0^{+}}(−x) = 0.

The second method works; the first one doesn’t.