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# L'Hospital's Rule L'Hospital's Rule is a method used to evaluate limits of indeterminate forms such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. **Theorem** Suppose that f and g are differentiable and $g'(x) \neq 0$ on an open interval I that contains a (except possibly at a), and that $\lim_{x...

# L'Hospital's Rule L'Hospital's Rule is a method used to evaluate limits of indeterminate forms such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. **Theorem** Suppose that f and g are differentiable and $g'(x) \neq 0$ on an open interval I that contains a (except possibly at a), and that $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ or that $\lim_{x \to a} f(x) = \pm \infty$ and $\lim_{x \to a} g(x) = \pm \infty$ Then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ if the limit on the right side exists (or is $\infty$ or $-\infty$). **Note** - L'Hospital's Rule says that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives, provided that the given conditions are satisfied. - It is especially important to verify the conditions regarding the limits of f and g before using L'Hospital's Rule. - L'Hospital's Rule is also valid for one-sided limits and for limits at infinity or negative infinity; that is, "x $\to$ a" can be replaced by any of the symbols "x $\to a^+$", "x $\to a^-$", "x $\to \infty$", or "x $\to -\infty$". - To use the L'Hospital's Rule the numerator and denominator must separately approach 0 or $\infty$. So if we have $\lim_{x \to a} \frac{f(x)}{g(x)}$ where $\lim_{x \to a} f(x) \neq 0$ and $\lim_{x \to a} g(x) = 0$,then the fraction grows without bound and does not approach $\frac{f'(x)}{g'(x)}$. **Indeterminate Products** If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = \infty$ (or $-\infty$), then it isn't clear what the value of $f(x)g(x)$, if any, will be. There is a struggle between f and g. If f wins, the answer will be 0; if g wins, the answer will be $\infty$ (or $-\infty$). Or there may be a compromise. To find out, we rewrite $f(x)g(x)$ as a quotient: $f(x)g(x) = \frac{f(x)}{1/g(x)}$ or $f(x)g(x) = \frac{g(x)}{1/f(x)}$ This converts the given limit into the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so that we can use L'Hospital's Rule. **Indeterminate Differences** If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, then the limit $\lim_{x \to a} [f(x) - g(x)]$ is called an indeterminate form of the type $\infty - \infty$. Again we can't guess the value of this limit until we have done some work on it. The general strategy is to rewrite the expression to convert it into either a quotient or a product. **Indeterminate Powers** Several indeterminate forms arise from the limit $\lim_{x \to a} [f(x)]^{g(x)}$ 1. $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ type $0^0$ 2. $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = 0$ type $\infty^0$ 3. $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \pm \infty$ type $1^\infty$ Each of these three cases can be treated either by taking the natural logarithm: let $y=[f(x)]^{g(x)}$, then $lny=g(x)lnf(x)$; or by writing the function as an exponential: $[f(x)]^{g(x)} = e^{g(x)lnf(x)}$. In either method we are led to the indeterminate product $g(x)lnf(x)$, which is of the type $0 \cdot \infty$.