L'Hospital's Rule and Indeterminate Forms

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Examples

Indeterminate Quotients $\displaystyle\frac{0}{0}$ or $\displaystyle\frac{\pm\infty}{\pm\infty}$

We start with indeterminate quotients, where either the numerator and denominator are both going to zero or are both going to infinity (positive or negative in any combination).

Example: Evaluate

$\displaystyle{\lim_{x \to 0} \frac{x}{1-e^x}}$ .

Solution: First, note that both

$f(x)=x$ and

$g(x)=1-e^x$ are zero when

$x=0$ . So L'Hospital's rule applies. Then

$\lim_{x \to 0} \frac{x}{1-e^x}$

$= \lim_{x\to 0} \frac{1}{-e^x} = \frac{1}{-e^0}$

$= \frac{1}{-1} = -1.$ Note that we are computing the ratio of

$f'(x)=1$ and

$g'(x)=-e^x$ , not the derivative of

$f/g$ !

Indeterminate Products $0 \cdot \pm\infty$

Here we consider indeterminate products where one factor is going to zero and the other is going to $\pm\infty$ . Notice: we cannot apply L'Hospital's rule to a product -- we must manipulate the product to get a quotient before applying the rule.