L'Hopital's Rule

  • Thread starter Nope
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  • #1
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Homework Statement


lim x->0
5x(cos 9x-1)/sin 5x-5x


Homework Equations





The Attempt at a Solution


answer is 243/25
The derivative of sin 5x-5x is always 0,
dunno know how to do it...
 

Answers and Replies

  • #2
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Is it sin(5x-5x) or sin(5x)-5x? The first one seems like a trivial way to express an equation and would result in 0 anyway, making the original equation undefined.
 
  • #3
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I'm not sure, it's an assignment from the internet,
it should be sin(5x-5x) , cause it ask you to use L'Hopital's Rule
sin(0)=0
 
  • #4
104
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Has to be

[tex]\lim_{x \to 0} \frac{5x(\cos(9x) - 1)}{\sin(5x) - 5x}[/tex]

OP the derivative of [tex]5x(\cos(9x) - 1)[/tex] is

[tex]5\cos(9x) - 5 - 45x \sin(9x)[/tex]

and the derivative of [tex]\sin(5x) - 5x[/tex] is

[tex]5\cos(5x) - 5[/tex]

Know how to proceed from here?
 
  • #5
16
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You'd use the rule if you got a 0 in the denominator after plugging in the limit of x, not if it were 0 to begin with. An equation with a 0 in the denominator would always have a 0 in the denominator no matter how many times you applied the rule.
 
Last edited:
  • #6
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You'd use the rule if you got a 0 in the denominator after plugging in the limit of x, not if it were 0 to begin with. An equation over 0 would always be over 0 no matter how many times you differentiate it.

Not quite...

An equation over zero is undefined.

x/0 is not equal to zero.

Edit: Actually I think you meant if you differentiate zero you get zero...
 
  • #7
16
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Edit: Actually I think you meant if you differentiate zero you get zero...

Yes, that's what I meant, but I see that I worded it incorrectly before. To explain a little better, though, I meant that if you applied the rule to an equation over 0 (f(x)/0), it would still be over 0 no matter how many times you differentiated the top and bottom.
 
  • #8
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[tex]
\lim_{x \to 0} \frac{5x(\cos(9x) - 1)}{\sin(5x) - 5x}
[/tex]
I tried
After the second derivative, the denominator is still zero...
-25sin(5x)
 
  • #9
16
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What's to stop you from differentiating a third time? Without the extra factor, it looks like the denominator would turn into a function of cosine, which wouldn't give you zero when you plug in zero.
 
  • #10
104
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Differentiate again and it will work.
 
  • #11
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ok ty, I will do it later, I have class now
 
  • #12
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I differentiated like 5 times, I still got 0..
I notice that there is always a x*cos(9x) in the nominator...which make whole thing equal to 0..
 
  • #13
HallsofIvy
Science Advisor
Homework Helper
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Yes, there is nothing at all wrong with getting "0" in the numerator. As long as the denominator is not also 0, that just tells you that the limit is 0.
 
  • #14
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I differentiated like 5 times, I still got 0..
I notice that there is always a x*cos(9x) in the nominator...which make whole thing equal to 0..

Then you've made a mistake somewhere, can get messy at times.

First derivatives.

[tex]\frac{5 \cos(9x) - 5 - 45x\sin(9x)}{5\cos(5x) - 5}[/tex]

Second one

[tex]\frac{90 \sin(9x) + 405x\cos(9x)}{25\sin(5x)}[/tex]

third one


[tex]\frac{1215 \cos(9x) + 3645x\sin(9x)}{125\cos(5x)}[/tex]

Now sub in x = 0
 

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