Computing Limit with L'Hospital's Rule: 1-e^(3x)/sinx

  • Thread starter sara_87
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In summary: It is common to taylor expand those kind of functions that become small so you can get something like this:(1-exp(3x))/sinx =(1-1-3x)/x +0(x^2)=-3.I don't think you can use de L'Hopital here because you have a pole.
  • #1
sara_87
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i have to use l'hospital's rule to compute the limit of:

lim (1-e^(3x))/sinx
x to 0

i got -3 bt I'm not sure

and what type is it? is it ''infinity/infinity'' type?
 
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  • #2
Looks like you applied L'Hopitals rule right to me - :)

When you say you're not sure what "type" it is, I'm assuming you mean what kind of indeterminate form? If you look at the function in your limit, actually evaluated at the limit value (x = 0), what kind of form is it? Surely not infinity over infinity... :)
 
  • #3
It is common to taylor expand those kind of functions that become small so you can get something like this:
(1-exp(3x))/sinx =(1-1-3x)/x +0(x^2)=-3.
I don't think you can use de L'Hopital here because you have a pole.
 
  • #4
Marco_84 said:
It is common to taylor expand those kind of functions that become small so you can get something like this:
(1-exp(3x))/sinx =(1-1-3x)/x +0(x^2)=-3.
I don't think you can use de L'Hopital here because you have a pole.

This is incorrect and confusing.

Taylor expansion and l'Hopital's rule are actually equivalent. And f/g does not have a pole at zero, because the limit is finite. 1/g has a pole, and that is precisely why you use l'Hopital's in the first place.
 
  • #5
Ben Niehoff said:
This is incorrect and confusing.

Taylor expansion and l'Hopital's rule are actually equivalent. And f/g does not have a pole at zero, because the limit is finite. 1/g has a pole, and that is precisely why you use l'Hopital's in the first place.

Great post =]
 
  • #6
retrofit81 said:
Looks like you applied L'Hopitals rule right to me - :)

When you say you're not sure what "type" it is, I'm assuming you mean what kind of indeterminate form? If you look at the function in your limit, actually evaluated at the limit value (x = 0), what kind of form is it? Surely not infinity over infinity... :)

oh ok so wold the type be ''0/0'' ?
 
  • #7
Yes =]
 

1. What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical theorem that helps to evaluate limits of functions that are in an indeterminate form, such as 0/0 or ∞/∞. It states that if the limit of the quotient of two functions is indeterminate, then taking the derivative of the numerator and denominator and re-evaluating the limit can help to determine the actual value.

2. How do you use L'Hospital's Rule to compute limits?

To use L'Hospital's Rule, first check if the limit of the function is in an indeterminate form. If it is, take the derivative of the numerator and denominator separately. Then, re-evaluate the limit using the new derivatives. If the limit is still in an indeterminate form, repeat the process until the limit can be evaluated.

3. What is the limit of 1-e^(3x)/sinx as x approaches 0?

The limit of 1-e^(3x)/sinx as x approaches 0 is 3. Using L'Hospital's Rule, we can take the derivatives of the numerator and denominator to get -3e^(3x)/cosx. Re-evaluating the limit with this new function gives us 3 as the final result.

4. Can L'Hospital's Rule be used for all types of functions?

No, L'Hospital's Rule can only be used for functions that are in an indeterminate form and the limit can be determined by taking the derivative. For example, it cannot be used for limits involving trigonometric functions like tanx or cotx.

5. Are there any limitations to using L'Hospital's Rule?

Yes, there are some limitations to using L'Hospital's Rule. It can only be used for limits that approach either 0 or ∞. Also, it may not always give the correct answer, so it is important to check the answer using other methods as well.

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