Question

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

Answer 1

Don't tell me what to do!

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

Oh, going to 0, whoops, not infinity x_x

Answer 6

yeah its going to 0

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

Okay I'll take the derivative again .

Answer 11

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.

Answer 12

so in the end the limit will just equal 0 ?

Answer 13

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

Answer 14

Yeah haha, okay thank you very much :)

Answer 15

yeah, np :3

Answer 16

@jossy04 Have you seen limits with indeterminate forms before?

Answer 17

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

Answer 18

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

Answer 19

I've used l'Hoptial's Rule before

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

@Ⓐ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

Answer 25

@jossy04

Answer 26

Ill butt out now, sorry.

Answer 27

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?

Answer 28

Thanks for all of your help @Psymon :)

Answer 29

1/0 ?

Answer 30

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

Answer 31

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

Answer 32

it dne ?

Answer 33

Are you using a calculator?

Answer 34

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

Answer 35

no i'm not using a calculator

Answer 36

Let's go back to your problem.

Answer 37

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

Answer 38

common denominators ?

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

its still 0 over 0

Answer 43

Exactly!

Answer 44

The original problem was not of this form.

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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.

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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