L'Hopital's Rule

" $0 / 0$ " is called an indeterminate form because knowing that $\mathrm{f}$ approaches $0$ and $\mathrm{g}$ approaches $0$ is not enough to determine the limit of $\mathrm{f} / \mathrm{g}$, even if it has a limit. The ratio of a "small" number divided by a "small" number can be almost anything as the three simple " $0/0$ " examples show:

$\lim \limits_{x \rightarrow 0} 3 x / x=3$, $\lim \limits_{x \rightarrow 0} x^{2} / x=0$, and  $\lim \limits_{x \rightarrow 0} 5 x / x^{3}=\infty$

Similarly, " $\infty/\infty$ " is an indeterminate form because knowing that $\mathrm{f}$ and $\mathrm{g}$ both grow arbitrarily large is not enough to determine the value limit of $\mathrm{f} / \mathrm{g}$ or if the limit exists:

$\lim \limits_{x \rightarrow \infty} 3 x / x=3$, $\lim \limits_{x \rightarrow \infty} x^{2} / x=\infty$, and $\lim \limits_{x \rightarrow \infty} 5 x / x^{3}=0$

Besides the indeterminate quotient forms " $0/0$ " and " $\infty/\infty$ " there are several other "indeterminate forms". In each case, the resulting limit depends not only on each function's limit but also on how quickly each function approaches its limit.

Product: If $f$ approaches $0$, and $g$ grows arbitrarily large, the product $\mathrm{f} \cdot \mathrm{g}$ has the indeterminant form " $0 \cdot \infty$ ".

Exponent: If $\mathrm{f}$ and $\mathrm{g}$ both approach $0$, the function $\mathrm{f}^{\mathrm{g}}$ has the indeterminant form " $0^{0}$ ".

If $f$ approaches $1$, and g grows arbitrarily large, the function $\mathrm{f}^{\mathrm{g}}$ has the indeterminant form " $1^{\infty}$ ".

If $f$ grows arbitrarily large, and $g$ approaches $0$, the function $f^{g}$ has the indeterminant form " $\infty^{0}$ ".

Difference: If $\mathrm{f}$ and $\mathrm{g}$ both grow arbitrarily large, the function $\mathrm{f}-\mathrm{g}$ has the indeterminant form " $\infty - \infty$ ".

Unfortunately, l'Hô pital's Rule can only be used directly with an indeterminate quotient (" $0/0$ " or " $\infty/\infty$ '), but these other forms can be algebraically manipulated into quotients, and then l'Hô pital's Rule can be applied to the resulting quotient.

Example 5: Evaluate $\lim \limits_{x \rightarrow 0^{+}} \mathrm{x} \cdot \ln (\mathrm{x})$ (" $0 \cdot(-\infty)$ " form)

Solution: This limit involves an indeterminate product, and we need a quotient in order to apply l'Hô pital's Rule. We can rewrite the product $x \cdot \ln (x)$ as the quotient $\frac{\ln (x)}{1 / x}$, and then so apply l'Hô pital's Rule

$\begin{aligned} \lim \limits_{x \rightarrow 0^{+}} x \ln (x) &=\lim \limits_{x \rightarrow 0^{+}} \frac{\ln (x)}{1 / x} \rightarrow \frac{\infty}{\infty} \\ &=\lim \limits_{x \rightarrow 0^{+}} \frac{1 / x}{-1 / x^{2}}=\lim \limits_{x \rightarrow 0^{+}}-x=0 \end{aligned}$

Example 6: Evaluate $\lim \limits_{x \rightarrow 0^{+}} \mathrm{x}^{\mathrm{x}} ( 0^{0}$  form)

Solution: An indeterminate exponent can be converted to a product by recalling a property of exponential and
logarithm functions: for any positive number $a, a=e^{\ln (a)}$ so $f^{g}=e^{\ln (f g)}=e^{g \ln (f)}$.
$\lim \limits_{x \rightarrow 0^{+}} x^{x}=\lim \limits_{x \rightarrow 0^{+}} e^{\ln \left(x^{x}\right)}=\lim \limits_{x \rightarrow 0^{+}} e^{x \cdot \ln (x)}$ and this last limit involves an indeterminate product $\mathrm{x} \cdot \ln (\mathrm{x}) \rightarrow 0 \cdot(-\infty)$ which we converted to a quotient and evaluated to be $0$ in Example 5.
Our final answer is then $\mathrm{e}^{\mathbf{0}}=\mathbf{1}$:

$\lim \limits_{x \rightarrow 0^{+}} x^{x}=\lim \limits_{x \rightarrow 0^{+}} e^{\ln \left(x^{x}\right)}=\lim \limits_{x \rightarrow 0^{+}} e^{x \cdot \ln (x)}=e^{0}=1$

Example 7: Evaluate $\lim \limits_{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}$ (" $1^{\infty}$ " form)

Solution: $\lim \limits_{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}=\lim \limits_{x \rightarrow \infty} \mathrm{e}^{\mathrm{x} \cdot \ln (1+\mathrm{a} / \mathrm{x})}$ so we need $\lim \limits_{x \rightarrow \infty} \mathrm{x} \cdot \ln \left(1+\frac{\mathrm{a}}{\mathrm{x}}\right)$ $\lim \limits_{x \rightarrow \infty} x \cdot \ln \left(1+\frac{a}{x}\right) \rightarrow$ " $\infty \cdot 0$ "  an indeterminate product so rewrite it as a quotient

$=\lim \limits_{x \rightarrow \infty} \frac{\ln \left(1+\frac{a}{x}\right)}{1 / x} \rightarrow  \frac{0}{0}$ an indeterminate quotient so use l'Hô pital's Rule

$=\lim \limits_{x \rightarrow \infty} \frac{\left(\frac{-a / x^{2}}{1+\frac{a}{x}}\right)}{\left(-\frac{1}{x^{2}}\right)}=\lim \limits_{x \rightarrow \infty} \frac{a}{1+\frac{a}{x}} \rightarrow \frac{\mathrm{a}}{1}=\mathbf{a}$

Finally, $\lim \limits_{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}=\lim \limits_{x \rightarrow \infty} e^{x \cdot \ln (1+a / x)}=\mathrm{e}^{\mathbf{a}}$.