l’Hôpital’s rule

L’Hôpital’s rule states that given an unresolvable limit of the form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$, the ratio of functions $\frac{f(x)}{g(x)}$ will have the same limit at $c$ as the ratio $\frac{f^{\prime }(x)}{g^{\prime }(x)}$. In short, if the limit of a ratio of functions approaches an indeterminate form, then

$$\underset{x\to c}{lim}\frac{f(x)}{g(x)}=\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

provided this last limit exists. L’Hôpital’s rule may be applied indefinitely as long as the conditions are satisfied. However it is important to note, that the nonexistance of $lim\frac{f^{\prime }(x)}{g^{\prime }(x)}$ does not prove the nonexistance of $lim\frac{f(x)}{g(x)}$.

Example:We try to determine the value of

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}.$$

As $x$ approaches $\mathrm{\infty }$ the expression becomes an indeterminate form $\frac{\mathrm{\infty }}{\mathrm{\infty }}$. By applying L’Hôpital’s rule twice we get

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{2x}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{2}{e^x}=0.$$

Another example of the usage of L’Hôpital’s rule can be found http://planetmath.org/node/5741here.