L’Hopital’s Rule Let’s start with some problems that involve taking limits. Problem: Evaluate each of the following limits. x 3 − 27 a. lim x→3 x − 3
x2 + 5 c. lim x→−∞ x + 3
b. lim
csc ( 3( x + h )) − csc ( 3x )
h→0
h
sin ( x ) x→0 x
d. lim
Each of those limits had to be dealt with in a slightly different way. For the (a), you might have been inclined to factor, cancel, and evaluate. For (b), you probably noticed it was a derivative right away— otherwise I’m not sure what you might have done. In (c), well, that’s just a nightmare problem to show the work for—actually, that one is never really going to get better. Finally, in (d)—I guess you either knew it or you didn’t, right? Well, we’re going to take care of that right now. Question: What did all of those limits have in common before you tried to use whatever technique you tried? [Hint: What should be the first step of trying to evaluate any limit?]
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There’s a rule we can apply to limits that fall into a certain category. Let’s see if we can figure it out. Problem: Suppose that f ( x ) and g ( x ) are both differentiable at x = a , but that lim x→a
f ( x) 0 = by direct g( x) 0
substitution. a. If f ( x ) and g ( x ) are differentiable at x = a , what else is true about them at that value? (Also, bonus question, what are the values of f ( a ) and g ( a ) ?)
b. Write the equation of the lines tangent to both f ( x ) and g ( x ) at x = a .
c. If a function has a tangent line at a point, then when you’re sufficiently close to that point, the function and the tangent line are virtually indistinguishable. They can practically be interchanged…which is totally what we’re going to do. Very, very close to x = a ,
f ( x) ≈
g( x) ≈
d. Since taking the limit as x approaches a involves being very, very close a, we can rewrite our limit:
lim x→a
f ( x) = g( x)
Which brings us to the rule. L’Hopital’s Rule f ( x) 0 ∞ If lim (or as x → ∞ ) yields or ± when you try to take the limit, then x→a g ( x ) 0 ∞ f ( x) f ′( x) . lim = lim x→a g ( x ) x→a g′ ( x ) 0 ∞ Note: You can keep repeating the process as long as you keep getting or ± . 0 ∞
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This is a hugely useful result. Don’t go crazy and over use it, though! You have to get an indeterminate 0 ∞ form (either or ± ), so always check. Also, it’s still more expedient to notice that some limits (in 0 ∞ fact, many limits) are actually just the definition of the derivative. Also, L’Hopital’s Rule is not the Quotient Rule! Problems: Evaluate each of the following limits. x 3 − 27 a. lim x→3 x − 3
c. lim x→1
e.
x2 − 1 x2 − x
cos ( x ) x→(π 2 ) 1− sin ( x ) lim +
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b. lim x→0
d. lim x→∞
f.
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sin ( x ) x
ln
( x) x2
sin ( 4x ) x→0 tan ( 5x )
lim
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g. lim+
ln ( x ) x
e x − 1− x h. lim x→0 x2
i.
lim
x − sin ( x ) x − tan ( x )
j.
lim
k. lim
x + sin ( x ) x + cos ( x )
l.
lim
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x→0
x→0
x→0
cos ( m ⋅ x ) − cos ( n ⋅ x ) x→0 x2
x→0
x tan
−1
( 4x )
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L’Hopital’s Rule is useful, but it can sometimes get stuck in a loop. Problem: Try to evaluate lim x→∞
x x2 + 1
using L’Hopital’s Rule.
Problem: Use another method to evaluate the limit above.
Problem: (Random) A laser pointer is placed on a rotating platform that turns at a rate of two revolutions per minute. The beam hits a wall that is 10 meters away, producing a beam that creates a red dot that moves horizontally along the wall. Let θ by the angle between the beam and the perpendicular line from the laser pointer to the wall. How fast is the dot moving when θ = π 3 ?
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Review(ish) Problems: Do them quickly! 2 3 1. Find all critical points of f ( x ) = ( 3x + 1) ( 7 − 2x ) .
2. Write the point-slope equation of the line tangent to the graph of y 2 − 2xy = 16 at ( 3,−2 ) .
3. Let f be the function given by f ( x ) = 5x 3 − 3x 2 + 6x − 4 . What is the instantaneous rate of change of f at x = −2 ?
1 4. A particle moves along the x-axis with position given by x ( t ) = t 3 − 5t 2 + 2t − 1 . Describe the 3 particle’s motion and position at t = 1 .
⎪⎧ 2x + a if x ≤ 1 f ( x) = ⎨ 3 if x > 1 ⎪⎩ bx 5. Let f ( x ) be the function given above. What are all values of a and b for which f ( x ) is differentiable at x = 1 ?
arctan ( a + h ) − arctan ( a ) 1 = , then what are all possible values of a? h→0 h 2
6. If lim
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