Question

Using l'hopitals rule please compute:

Answer 1

\[\lim_{n \rightarrow \infty} n^2(1-\cos(\frac{ 5.2 }{ n}))\]

Answer 2

@tkhunny

Answer 3

@ikram002p

Answer 4

For starters, we need to get it into a form where the rule applies. Which indeterminate form are we looking at?

Answer 5

It doesn't mention in the question. Can you give me the easiest one?

Answer 6

?? It's not a worked example. It's a think about it problem. Can we use l'hospital or not? It has to be an indeterminate form or it's no good.

Answer 7

Yeah ofcourse we can use l'hopitals if we divide by n which would make it n^-2

Answer 8

You didn't quite say that right, but it seems you have the right idea. Good work.

So, now we have this: \(\dfrac{1 - \cos\left(\dfrac{5.2}{n}\right)}{\dfrac{1}{n^{2}}}\)

What indetermiante form is that?

Answer 9

0/inf

Answer 10

sorry thats 0/0

Answer 11

Since the first one is no good, I am delighted that you fixed it. Okay, now for the derivatives.

Answer 12

the top would be sin(5.2/n) (-5.2n^-2) all divided by -2n^-3

Answer 13

@tkhunny are you there?

Answer 14

Its here where I got stuck and needed help.

Answer 15

Yes. I'm waiting for you to simplify your result. That's horrid.

Answer 16

I can't do it, I got stuck over here.

Answer 17

*eats popcorn* This is great.
Hint: Get rid of the denominator!

Answer 18

I actually don't believe that. Did you take algebra or not? I'm going with "yes".

The numerator has n^(-2) and the denominator has n^(-3). Simplify that pair.

Answer 19

Got stuck on the easiest part mate lol.

Answer 20

thats n^(-1) on bottom which is equivalent to n on the numerator. I have been doing all types of integrals all day! Sorry guys <3.

Answer 21

Well, take a deep breath and do one more. Keep doing it until you get \(2.6n\sin(5.2/n)\)

Answer 22

After that what do I do? The answer isn't infinity, its 13.52. So I take the derivative again?

Answer 23

So jumpy. Did you take that deep breath?

You MAY need another derivative. Did anyone tell you that l'hospotal's rule doesn't always work, even if it is an indeterminate form?

Anyway, start over. Is this expression an indeterminate form? Just like we started - think it over at each step. As soon as it's NOT in indeterminate form, we have to think of something else. So, it it? CAN we do another derivative?

Answer 24

Yeah, use the product rule. to find the derivative. I know that you can take the limit inside the function but I don't know when to do that, when we have got the right form?

Answer 25

?? That doesn't make any sense. Did you look at this form? \(\dfrac{2.6\cdot\sin(5.2/n)}{1/n}\)

Is that an indeterminate form?

Answer 26

Yes this is inf/0

Answer 27

That is not correct. Look at the numerator a little harder. Remember, it's sine up there, not cosine.

Answer 28

0/0 then!

Answer 29

@iPwnBunnies are you still eating popcorn mate?

Answer 30

Yep. Now, apply L'Hopital's Rule again.

Answer 31

Please do.

Answer 32

2.6n = u and sin(5.2/n) = v
u' = 2.6 and v' =cos(5.2/n)(1/n)

Product rule = u'v + v'u
2.6sin(5.2/n) + cos(5.2/n)(1/n)(2.6n)

Answer 33

There is no product rule. The variable appears only once in both numerator and denominator.

Answer 34

I took the n back up and included it with the 2.6. It is equivalent to what you wrote.

Answer 35

l'hospital's rule does not tell you to find the derivative of the entire expression. You cannot do it. You must find the indeterminate form and then treat numerator and denominator separately.

Answer 36

Oh okay! Then here is the part that I went wrong in.

So it would be 2.6cos(5.2/n)(1/n) all divided by -1/n^2

Answer 37

@tkhunny am I right or did I do a silly mistake again?

Answer 38

There are no silly mistakes when they cause you to learn.
Did you make any deliberate errors?

Try the numerator again.

Answer 39

@tkhunny No but its late night and I have been studying all day I can't think anymore. Anyways I got your idea, thank you very much for your help! I really appreciate it.

Answer 40

If you are careful, you should get \(13.52\cos(5.2/n)\) and that should be helpful.

Answer 41

Aww man, you're almost there. You messed up the chain rule in the numerator.

Answer 42

@tkhunny thank you very much for your help!

Answer 43

@iPwnBunnies Thank you very much for your sensitive emotions.