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

[sara_87]

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?

 

[retrofit81]

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... :)

 

[Marco_84]

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.

 

[Ben Niehoff]

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.

 

[Gib Z]

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 =]

 

[sara_87]

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'' ?

 

[Gib Z]

Yes =]