True or False? A limit that L'Hospital's rule can evaluate can often be evaluated another way as well.

True. The rule is one route among several rather than the only one. Where substitution gives an indeterminate form, algebraic simplification, multiplying by a conjugate and multiplying by a reciprocal are all still available.

What do you do when L'Hospital's rule leaves you with another indeterminate form?

Apply the rule again to the new quotient, differentiating top and bottom once more. That can be repeated as often as needed, working through higher and higher derivatives, until the form is no longer indeterminate.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 4: Contextual Applications of DifferentiationL'Hospital's Rule

L'Hospital's Rule (College Board AP® Calculus BC): Flashcards

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FrontL'Hospital's Rule
Define an indeterminate form.

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Cards in this collection (9)

Define an indeterminate form.
An indeterminate form is an expression reached by substitution that does not tell you the value of the limit.
The two to know are $\frac{0}{0}$ and $\frac{\pm \infty}{\pm \infty}$ .
Why is $\frac{\infty}{\infty}$ not equal to $1$ ?
Because $\infty$ is not a number, so it cannot be cancelled the way a common factor in a fraction can.
It is a label saying that the limit is still unknown, which is also why $\lim_{x \to a} f (x) = \frac{0}{0}$ should never be written.
True or False?
L'Hospital's rule tells you to differentiate $\frac{f (x)}{g (x)}$ using the quotient rule.
False.
The rule replaces the quotient by the ratio of the two derivatives, $\frac{f^{'} (x)}{g^{'} (x)}$ .
Reaching for the quotient rule here is a reliable sign that something has gone wrong.
Fill in the two missing derivatives in L'Hospital's rule:
$\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{______}{______}$
The completed rule is $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ .
Top and bottom are differentiated separately, and the limit is then attempted again in the new form.
What must you check before applying L'Hospital's rule?
That substitution really does give one of the two indeterminate forms, so that top and bottom both tend to $0$ or both tend to $\pm \infty$ .
Without that the rule is simply not valid, whatever answer it happens to produce.
True or False?
A limit that L'Hospital's rule can evaluate can often be evaluated another way as well.
True.
The rule is one route among several rather than the only one.
Where substitution gives an indeterminate form, algebraic simplification, multiplying by a conjugate and multiplying by a reciprocal are all still available.
Use L'Hospital's rule to evaluate $\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x}$ .
It is $- \frac{5}{3}$ .
Substitution gives $\frac{\infty}{- \infty}$ , and differentiating top and bottom leaves $\lim_{x \to \infty} \frac{5}{- 3}$ , which has no $x$ in it at all.
Substitution gives $\frac{0}{\pm \infty}$ for one limit and $\frac{\pm \infty}{0}$ for another. Can L'Hospital's rule be used?
No, because neither of those is an indeterminate form.
The first limit is simply $0$ , and the second diverges to $\infty$ or $- \infty$ depending on the signs near the point.
What do you do when L'Hospital's rule leaves you with another indeterminate form?
Apply the rule again to the new quotient, differentiating top and bottom once more.
That can be repeated as often as needed, working through higher and higher derivatives, until the form is no longer indeterminate.

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