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L’Hospital’s Rule. I’m the real L’Hopital! And they’re both bad spellers. ^. To Tell the Truth. My name is L’Hospital! And that first guy can’t spell. Hello, my name is L’Hopital. L’Hospital #1. L’Hospital #3. L’Hospital #2. Question: Messieurs, can you tell us
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I’m the real L’Hopital! And they’re both bad spellers. ^ To Tell the Truth My name is L’Hospital! And that first guy can’t spell. Hello, my name is L’Hopital.
L’Hospital #1 L’Hospital #3 L’Hospital #2 Question: Messieurs, can you tell us something about your famous rule?
Will the real L’Hospital please stand up!!! L’Hospital #3 Johann Bernoulli It’s true, L’Hospital’s Rule can be directly applied to the limit forms and . But the rule says nothing if doesn’t exist. L’Hospital #1 L’Hospital #3 L’Hospital #2
Let a be a real number or and I an open interval which contains a or has a as an endpoint. or b a c [Suppose that and for all x in I.] This condition is typical, but not needed if we assume exists. If or , and , then . Here’s a correct statement of L’Hospital’s Rule: This just takes care of one-sided limits and limits at infinity all at once.
Let f and g be continuous on and differentiable on . Then there is a c in with or . The many different cases of L’Hospital’s Rule can all be proven using Cauchy’s Mean Value Theorem: Extra letters and limits of rachieauxs seem to be French things. Maybe it’s something in the Perrier. Eau well!
Apply Rolle’s Theorem to the function on the interval . Cauchy’s Mean Value Theorem can be proven from Rolle’s Theorem: See Handout!
The case is covered in most textbooks, but the case isn’t mentioned. From Cauchy, we get that
Case #1: If and are sufficiently large, then we can make and arbitrarily close to zero, since , and arbitrarily close to . So we get that .
Case #2: If and are sufficiently close to a, then we can make and arbitrarily close to zero, since , and arbitrarily close to . So we get that .
1. 2. Examples where L’Hospital’s Rule doesn’t apply: See Handout! See Handout!
3. 4. Examples where L’Hospital’s Rule doesn’t apply(cont.): See Handout! See Handout!
5. An example where you can’t get away from the zeros of . Examples where L’Hospital’s Rule doesn’t apply(cont.): See Handout!
1. 2. Watch out for !!! Surprising examples where L’Hospital’s Rule applies: See Handout! See Handout!
3. If exists on and , then find . If L is a number, then find . {Hint: Apply L’Hospital’s Rule to , and then observe that .} What’s if ? Is this a problem? More surprising examples where L’Hospital’s Rule applies: See Handout!
4. If exists on and , and exists as a number, then what must be the value of L? {Hint: Apply L’Hospital’s Rule to , and use it to determine . Determine the limit from the fact that exists as a number .} More surprising examples where L’Hospital’s Rule applies: See Handout!
5. If exists on and , where A is a number. Show that there is a sequence with and . {Hint: The Mean Value Theorem implies that for each n for some . So .} You might think that , by applying L’Hospital’s Rule in reverse, but consider . See Handout!
6. You can see that doesn’t exist. If we write the limit as and try L’Hospital’s Rule, we get . What’s wrong? See Handout!
Hint: 7. Use L’Hospital’s Rule to evaluate Where the numbers are arbitrary real numbers. See Handout!
A common test for determining the nature of critical numbers in a first semester Calculus course is the 2nd Derivative Test. Here is a list of three common hypotheses from six Calculus textbooks: I. Suppose that is continuous in an open interval containing c. II. Suppose that exists in an open interval containing c. III. Suppose that exists in an open interval containing c, and exists. The weakest hypothesis is III.
The following conclusions are common to all versions of the 2nd Derivative Test: If and , then f has a local maximum at . If and , then f has a local minimum at . If and , then the 2nd Derivative Test fails.
0 0 c c Suppose that exists in an open interval containing c, exists, and . See Handout! If , then the sign chart of looks like: If , then the sign chart of looks like:
If you ask a veteran Calculus student about the 2nd Derivative Test, you’ll probably get a positive response, but if you ask about the Nth Derivative Test, you’re likely to get a puzzled look. Typically, the Nth Derivative Test is proved using Taylor’s Theorem along with the following hypotheses in second semester Calculus or higher: For , suppose thatare continuous in an open interval containing and that , but .
We can prove an Off-the-rack Nth Derivative Test (without using Taylor’s Theorem) and with weaker hypotheses, i. e. first semester Calculus style. First, let’s find some general hypotheses on and its derivatives. Beginning of the Off-the-rack Nth Derivative Test: For , suppose that exist in an open interval containing , , exists and . Now we’ll investigate four cases:
Case I: is odd and : - - + + 0 c + + + + 0 c + + + + 0 c The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following: Odd derivative: Even derivative: Odd derivative:
Case II: is odd and : - - + + 0 c - - - - 0 c - - - - 0 c The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following: Odd derivative: Even derivative: Odd derivative:
Case III: is even and : + + + + 0 c - - + + 0 c The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following: Even derivative: - - + + 0 Odd derivative: c Odd derivative:
Case IV: is even and : - - - - 0 c + + - - 0 c The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following: Even derivative: - - + + 0 Odd derivative: c Odd derivative:
From the sign patterns in the previous four cases, we can now state an Off-the-rack Nth Derivative Test: For , suppose that exist in an open interval containing , , exists and . If is odd, then f has neither a maximum or minimum at . If is even and , then f has a local minimum at . If is even and , then f has a local maximum at .
Determine what’s going on at zero for the following functions:
Consider the functions: Suppose that f has derivatives of all orders in an open interval containing and they’re all equal to zero at . Can we conclude anything about the nature of f at ?
- - + + 0 c - - + + 0 c In the case of n being odd, can we conclude anything about the graph of f at x = c? See Handout! In the case of , we can conclude that the sign chart for f” near x = c is as follows: In the case of , we can conclude that the sign chart for f” near x = c is as follows: The Nth Derivative Test is fairly definitive.
Suppose that g has a (piecewise)continuous derivative on the interval and on . By considering the formula for the length of the graph of g on the interval , Determine the maximum possible length of the graph of g on the interval . Determine the minimum possible length of the graph of g on the interval . See Handout!
If , then complete the graph of the function g on the interval that has the maximum length. See Handout!
If , then complete the graph of the function g on the interval that has the minimum length. See Handout!
Do the same under the assumptions: or See Handout!
Suppose that f and g have (piecewise)continuous derivatives, , , and , then use the surface area of revolution about the y-axis formula to find a decent upper bound on the surface area of revolution about the y-axis of the curve See Handout!
to find the minimal surface area of revolution about the y-axis of the curve See Handout!
Fermat’s problem: Given three points in the plane, find a fourth such that the sum of its distances to the three given ones is a minimum.” Euclidean Steiner tree problem: Given N points in the plane, it is required to connect them by lines of minimal total length in such a way that any two points may be interconnected by line segments either directly or via other points and line segments. The Euclidean Steiner tree problem is solved by finding a minimal length tree that spans a set of vertices in the plane while allowing for the addition of auxiliary vertices (Steiner vertices). The Euclidean Steiner tree problem has long roots that date back to the 17th century when the famous scientist Pierre Fermat proposed the following problem: Find in the plane a point, the sum of whose distances from three given points is minimal.
A practical example: Two factories are located at the coordinates and with their power supply located at , . Find x so that the total length of power line from the power supply to the factories is a minimum. power supply Steiner point or vertex factory factory
The length of the power line as a function of x with the parameters a and h is given by
Here are the possible sign charts for L’ depending on the values of the parameters a and h.
So in the case of the Steiner point should be located units above the factories. If , then the power lines should go directly from the factories to the power supply without a Steiner point.
Compare this with the soap film configuration using the frames.