anonymous
  • anonymous
Use l'Hopital's Rule The limit as x approaches 0 of ((1/sinx)-(1/x))
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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abb0t
  • abb0t
Don't tell me what to do!
Psymon
  • 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
  • anonymous
so i would end up with x-sinx/xsinx ?

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Psymon
  • Psymon
Yeah, the derivatives run into an infinite loop, not immediately sure how to handle that xD
Psymon
  • Psymon
Oh, going to 0, whoops, not infinity x_x
anonymous
  • anonymous
yeah its going to 0
Psymon
  • 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
  • Psymon
Otherwise I'm not sure what else, Im probably missing an identity xD
Psymon
  • Psymon
After two applications of l'hopitals rule Im able to get it to be a defined answer.
anonymous
  • anonymous
Okay I'll take the derivative again .
Psymon
  • 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
  • anonymous
so in the end the limit will just equal 0 ?
Psymon
  • Psymon
Pretty unsatisfying, huh? But yes, thats what I get.
anonymous
  • anonymous
Yeah haha, okay thank you very much :)
Psymon
  • Psymon
yeah, np :3
anonymous
  • anonymous
@jossy04 Have you seen limits with indeterminate forms before?
anonymous
  • anonymous
Have you used l'Hopital's Rule before? Or is it the first time you've seen it?
anonymous
  • 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
  • anonymous
I've used l'Hoptial's Rule before
anonymous
  • anonymous
Ok good, I can help you with this one then. Usually, you use l'Hopital's Rule with fractions, right?
Psymon
  • 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
  • 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
  • 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
  • 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
  • anonymous
@jossy04
Psymon
  • Psymon
Ill butt out now, sorry.
anonymous
  • 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
  • anonymous
Thanks for all of your help @Psymon :)
anonymous
  • anonymous
1/0 ?
anonymous
  • anonymous
Right, but what does the zero in the denominator mean? You can never divide by zero.
anonymous
  • anonymous
What happens to 1/x as x gets really close to zero?
anonymous
  • anonymous
it dne ?
anonymous
  • anonymous
Are you using a calculator?
anonymous
  • anonymous
1/0 does not exist, but the limit of 1/x as x goes to zero does have an answer.
anonymous
  • anonymous
no i'm not using a calculator
anonymous
  • anonymous
Let's go back to your problem.
anonymous
  • 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
  • anonymous
common denominators ?
anonymous
  • anonymous
yes, Good. Let's try combining these using a common denominator.
anonymous
  • anonymous
on the top you would get x-sinx and on the bottom it would be xsinx
anonymous
  • anonymous
Good. Now let's look at what happens to both of these as x goes to zero.
anonymous
  • anonymous
its still 0 over 0
anonymous
  • anonymous
Exactly!
anonymous
  • anonymous
The original problem was not of this form.
anonymous
  • anonymous
so after that do we use l'hopital's rule again ?
anonymous
  • anonymous
If you get 0/0 or infinity over infinity, then (and only then) you can use l'Hopital's Rule.
anonymous
  • anonymous
Now you can use l'Hopital...do you know what l'Hopital's Rule says?
anonymous
  • 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
  • anonymous
What do you have to do to use l'Hopital's Rule?
anonymous
  • anonymous
Have you seen problems of the form 0/0 before?
anonymous
  • anonymous
yes i've seen problems with the form 0/0
anonymous
  • anonymous
What do you have to do to the numerator and denominator to find the limit?
anonymous
  • anonymous
What does l'Hopital's Rule say? @jossy04
hartnn
  • 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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