L'Hospital's Rule

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Hey I'm doing indeterminate products and using l'hospital's rule and does anyone have a strategy that works well in determining which function you put as your denominator? Obviously there isn't one strategy that works 100% but I'm just wondering if anyone knows a good way of determining this? Thanks.
 

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  • #2
SammyS
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Hey I'm doing indeterminate products and using l'hospital's rule and does anyone have a strategy that works well in determining which function you put as your denominator? Obviously there isn't one strategy that works 100% but I'm just wondering if anyone knows a good way of determining this? Thanks.
You're right, there is no general rule.

Whatever gives better cancellation of offending factors is often a good choice.
 
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What's a good way to start then? Putting the function whose derivative is simpler on the bottom or top or does it even matter?
 
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Why don't you provide us with an problem that you are having difficulty with. As you mentioned, there is no one way of getting your indeterminate form fit 0/0 or infinty/infinity. So, giving us an example may allow us to help you.
 
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SammyS
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What's a good way to start then? Putting the function whose derivative is simpler on the bottom or top or does it even matter?
It's not always that simple, (Pardon the joke.) although that does often work.

I have seen cases where the derivative of top or bottom gets more complicated than the original, but the result affords just the right amount of cancelling with the other derivative so that the overall result is nicely behaved.
--- no example off the top of my head.​
 
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Why don't you provide us with an problem that you are having difficulty with. As you mentioned, there is no one way of getting your indeterminate form fit 0/0 or infinty/infinity. So, giving us an example may allow us to help you.

Alright I think I did this one correctly

lim x--> 1+ ln(x) * tan(∏x/2)

I found this limit by putting ln(x) in the numerator and tan(∏x/2) in the denominator. I used l'hospital's rule once and found it to be 1/∞ so my answer is 0. Is this correct?
 
  • #7
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You can only use L'Hospital's Rule when it is 0/0 or infinity/infinity. If it is 0/infinity, you have to do some algebra to rewrite it in a form that is 0/0 or infinity/infinity.
 
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SammyS
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Alright I think I did this one correctly

lim x--> 1+ ln(x) * tan(∏x/2)

I found this limit by putting ln(x) in the numerator and tan(∏x/2) in the denominator. I used l'hospital's rule once and found it to be 1/∞ so my answer is 0. Is this correct?
Please show your steps.
 
  • #9
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What's the statement of the original problem? Is it [itex]lim_{x \rightarrow 1^+}~ ln(x) \cdot \tan( \frac{\pi x}{2})[/itex]? I ask, because you can't just arbitrarily chose which function you want to put in the numerator and which one will go in the denominator. You have to use algebraic manipulation to put it in the form you want.
 
  • #10
Dick
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Alright I think I did this one correctly

lim x--> 1+ ln(x) * tan(∏x/2)

I found this limit by putting ln(x) in the numerator and tan(∏x/2) in the denominator. I used l'hospital's rule once and found it to be 1/∞ so my answer is 0. Is this correct?

No, it's not right. How exactly did you do that? You don't just 'put something into the denominator'. a*b is not equal to a/b, it is equal to a/(1/b).
 
  • #11
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It appears we are all giving the same advice. @maceng7 is what we are saying making any sense?

Let a = ln(x) and b = tan(pix/2). As dick mentioned, a*b is entirely different from a/b. By putting the b in the denominator, you've created a new function, entirely different from the original; and when you evaluate the limit of a/b, the result you get tells you nothing of the limit for a*b. This is true, because a*b and a/b are different functions.
 
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  • #12
Ray Vickson
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You can only use L'Hospital's Rule when it is 0/0 or infinity/infinity. If it is 0/infinity, you have to do some algebra to rewrite it in a form that is 0/0 or infinity/infinity.

NO: if it is 0/∞ you are done: the answer = 0 and there is no need for l'Hospital's rule.
 
  • #13
SammyS
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NO: if it is 0/∞ you are done: the answer = 0 and there is no need for l'Hospital's rule.
But as Dick point's out, the limit is not zero so ...


Let's wait to hear back from OP .
 

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