How Do You Solve the Limit of (e^x * tan^-1 y) / y as (x, y) Approaches (0,0)?

[schattenjaeger]

Assuming it exists(and upon checking the answer is 1, but I'll be damned if I can get it)

(e^x * tan^-1 y) / y

the limit as (x,y)->(0,0)

I can't find a path where the denominator isn't 0 or cancels out that goes through the point 0,0, so I can't, you know, solve it.

 

[Berislav]

What is the Gauss aproximation for tan^-1 for small angles? e^0=1. Not the most exact way, but it should work.

 

[schattenjaeger]

Well, I'm in Calculus 3 and have no clue what that approxamation is nor do I think I'm expected to for this problem, thanks though

 

[vsage]

See next post.

 

[schattenjaeger]

Well, it's a function of two variables, and this problem is in the chapter before partial derivatives, so ignoring the fact that I'm not technically sposed to know how to apply l'hopital's rule to it, I don't even know if you can on a multivariable thing like that

 

[vsage]

Well, it's a function of two variables, and this problem is in the chapter before partial derivatives, so ignoring the fact that I'm not technically sposed to know how to apply l'hopital's rule to it, I don't even know if you can on a multivariable thing like that

Alright that puts the question in a little perspective. Still, you can simplify the limit to one variable by just applying x = 0 to it.

Edit: Applying tan^-1() to each side of the first approximation Berislav gave would work also I guess.

 

[Berislav]

Gauss aproximation:

siny=y, for very small y --->sin^-1y=y

cosy=sqrt(1-y^2)

tany=y/(sqrt(1-y^2))

The y's will cancel when you combine the above eqaution with your own.

 

[schattenjaeger]

Then I have to take the derivative of tan^-1y, which definitely isn't 1(like it would have to be for l'hopital's rule to give me 1)

As for the Gauss approximation, I don't think I should use it here because it doesn't specify only small values of y, and I certainly haven't officially learned it yet

 

[Hurkyl]

For small values of y, sin y = y, cos y = 1, and tan y = y.

 

[vsage]

Then I have to take the derivative of tan^-1y, which definitely isn't 1(like it would have to be for l'hopital's rule to give me 1)

How's it not? $$\frac{d(tan^{-1}(y))}{dy} = \frac{1}{1+y^2}$$ at y = 0.. Divide this by the derivative of the denominator of your original limit.

 

[Berislav]

For small values of y, sin y = y, cos y = 1, and tan y = y.

Right. I only approximated sin y=y, though.

 

[schattenjaeger]

Oh right

ok, thanks!