Tricky limit

[jakeisthesnake]

Hello Folks!

I am working on my asking-for-help technique, so let me know if the following problem isn't clear.

Any ideas for how to take the limit as t goes to 0?

Thanks! If you want to know why I am trying to solve this, see my longer post.
Jake

 

[jakeisthesnake]

progress

So I made some progress! The denominator simplifies to:

I isolated all of the terms with sin(t), so now I am trying to solve lim as t -> 0:

any tips for here?

Thanks as always,
Jake

 

[ksdhart]

I don't think there's anything that can be done to make it spectacularly easy. You just have to simplify every expression you can, and I think at some point you'll need to use L'Hopital's Rule. You say you're interested in the limit as t approaches 0, but make no mention of any of the other variables approaching a value, so then they're all treated as constants. Furthermore, since it's just four (admittedly very complex) terms being subtracted, I'd be inclined to tackle each one separately. Consider the numerator of the first fraction:

\(\displaystyle R\left(cos\left(\theta \right)cos\left(t\right)-sin\left(t\right)sin\left(\theta \right)\right)+R\left(\theta +t\right)\left(sin\left(\theta \right)cos\left(t\right)+sin\left(t\right)cos\left(\theta \right)\right)-A\left(sin\left(\theta \right)cos\left(t\right)+sin\left(t\right)cos\left(\theta \right)\right)\)

After applying angle addition identities, that mess simplifies to:

\(\displaystyle R\left(cos\left(\theta +t\right)\right)+R\left(\theta +t\right)\left(sin\left(\theta +t\right)\right)-A\left(sin\left(\theta +t\right)\right)\)

Then play with the denominator:

\(\displaystyle \frac{sin\left(\phi -\theta \right)}{cos\left(\phi -\theta \right)}-\frac{sin\left(\phi -\theta -t\right)}{cos\left(\phi -\theta -t\right)}=tan\left(\phi -\theta \right)-tan\left(\phi -\theta -t\right)\)

Apply the following identity:

\(\displaystyle tan\left(a\right)-tan\left(b\right)=sec\left(a\right)\cdot sec\left(b\right)\cdot sin\left(a-b\right)\)

And you can simplify the denominator down to:

\(\displaystyle sec\left(\phi -\theta \right)\cdot sec\left(\phi -\theta -t\right)\cdot sin\left(t\right)\)

Now, use L'Hopital's Rule on the fraction, and see what you get. Then tackle the other three fractions in a similar manner. I'd make a special note that the numerator of the fourth fraction has no t terms anywhere, so that makes things even easier. Good luck.

 

[Ishuda]

Use ksdhart's simplification and simplify a bit further:
Numerator:
N = \(\displaystyle cos\left(\phi -\theta \right)\, cos\left(\phi -\theta -t\right)\, [R\, cos\left(\theta +t\right)\, +\, (\, R\left(\theta +t\right)\, -\, A)\, sin\left(\theta +t\right)]\)
Denominator:
D = \(\displaystyle sin\left(t\right)\)

Now take the individual limits as t goes to zero:
N \(\displaystyle \to\, cos^2\left(\phi -\theta \right)\, [R\, cos\left(\theta\right)\, +\, \left(R\, \theta\, -\, A\right)\, sin\left(\theta\right)]\)

D \(\displaystyle \to\) 0.

So, unless there is something in the relationships between the other items [R, A, \(\displaystyle \theta\) and \(\displaystyle \phi\)] the limit is infinity.

 

[jakeisthesnake]

Thank you ksdhart and Ishuda!

ksdhart - Yeah, it looks like I will have to do some l'hopitaling. I hadn't considered that in the angle addition form so that was a great idea.

Ishuda - I think that I will have to take the other terms of the fraction into consideration because the final answer should be some non-zero, non-infinite function of theta. There is a chance that the steps that lead to taking this limit have a error, but I have checked several times, so I don't think so. (there is a link to my original post that describes the larger problem that I am working on.)

I'll post back here with any more progress (hopefully this weekend.)

Thanks again,
Jake