Use l'Hopital's Rule
The limit as x approaches 0 of ((1/sinx)-(1/x))

- anonymous

Use l'Hopital's Rule
The limit as x approaches 0 of ((1/sinx)-(1/x))

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- abb0t

Don't tell me what to do!

- Psymon

Well we want this to be put into an indeterminant form first. So I would use common denominators and make what you have into one fraction.

- anonymous

so i would end up with x-sinx/xsinx ?

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## More answers

- Psymon

Yeah, the derivatives run into an infinite loop, not immediately sure how to handle that xD

- Psymon

Oh, going to 0, whoops, not infinity x_x

- anonymous

yeah its going to 0

- Psymon

Well, what we would need to do is take derivatives until we do not end up in an indeterminant form. This will occur once we have a cos(x) in the denominator. That way no matter what we get int he numerator or for other portions of the denominator, we will have a defined answer.

- Psymon

Otherwise I'm not sure what else, Im probably missing an identity xD

- Psymon

After two applications of l'hopitals rule Im able to get it to be a defined answer.

- anonymous

Okay I'll take the derivative again .

- Psymon

Mhm. You should be able to get a cos(x) portion in the bottom which will allow us to prevent the dreaded 0 in the denominator.

- anonymous

so in the end the limit will just equal 0 ?

- Psymon

Pretty unsatisfying, huh? But yes, thats what I get.

- anonymous

Yeah haha, okay thank you very much :)

- Psymon

yeah, np :3

- anonymous

@jossy04 Have you seen limits with indeterminate forms before?

- anonymous

Have you used l'Hopital's Rule before?
Or is it the first time you've seen it?

- anonymous

Yeah , I have we barely started learning them, so its still a bit confusing for me. I'm kind of getting the hang of it though

- anonymous

I've used l'Hoptial's Rule before

- anonymous

Ok good, I can help you with this one then.
Usually, you use l'Hopital's Rule with fractions, right?

- Psymon

Basically the idea is you just need to make sure you get your function into an indeterminant form. If it doesnt start in an indeterminant form you need a way to make it into one. There may be a sneaky trick here and there to achieve that, but once you are able to you seem to know the process. Derivative of the numerator divided by the derivative of the denominator.

- anonymous

If the numerator and denominator both go to infinity, you can apply l'Hopital's Rule? Have you seen examples like that? @jossy04

- Psymon

Although it sounds like Archie may have a different answer than what I got, so I suppose she'll rework you through it. One last thing I'd like to add, though. I want to show you this trick. Say we have a function like:
\[\lim_{x \rightarrow \infty}xln(x)\] Now we don't have an indeterminant form really here. But this is one trick to be aware of. I can rewrite this function:
\[\lim_{x \rightarrow \infty}\frac{ \ln(x) }{ 1/x }\] Now we can apply l'hopitals with this.
Okay, im done interrupting, sorry xDD

- anonymous

@ⒶArchie☁✪ Yes I've seen examples of that.
@Psymon I get that , but I just don't get how you know how to see which indeterminate form it is

- anonymous

@jossy04

- Psymon

Ill butt out now, sorry.

- anonymous

our problem has two parts:
1/sin(x) and 1/x Do you know what the limit is for each of these as x goes to zero?

- anonymous

Thanks for all of your help @Psymon :)

- anonymous

1/0 ?

- anonymous

Right, but what does the zero in the denominator mean?
You can never divide by zero.

- anonymous

What happens to 1/x as x gets really close to zero?

- anonymous

it dne ?

- anonymous

Are you using a calculator?

- anonymous

1/0 does not exist, but the limit of 1/x as x goes to zero does have an answer.

- anonymous

no i'm not using a calculator

- anonymous

Let's go back to your problem.

- anonymous

l'Hopital's rule is usually applied to a fraction. But you have two fractions. How can you combine these into one? @jossy04

- anonymous

common denominators ?

- anonymous

yes, Good. Let's try combining these using a common denominator.

- anonymous

on the top you would get x-sinx and on the bottom it would be xsinx

- anonymous

Good. Now let's look at what happens to both of these as x goes to zero.

- anonymous

its still 0 over 0

- anonymous

Exactly!

- anonymous

The original problem was not of this form.

- anonymous

so after that do we use l'hopital's rule again ?

- anonymous

If you get 0/0 or infinity over infinity, then (and only then) you can use l'Hopital's Rule.

- anonymous

Now you can use l'Hopital...do you know what l'Hopital's Rule says?

- anonymous

Yes. But we have not really used it yet in this problem. First we had to rewrite this problem to get it into the form 0/0.

- anonymous

What do you have to do to use l'Hopital's Rule?

- anonymous

Have you seen problems of the form 0/0 before?

- anonymous

yes i've seen problems with the form 0/0

- anonymous

What do you have to do to the numerator and denominator to find the limit?

- anonymous

What does l'Hopital's Rule say? @jossy04

- hartnn

i'd like to share my thoughts,
try
(x csc x -1) \x (can we apply LH here ?) (0/0 form ?)

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