- #1

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(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.

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- Thread starter schattenjaeger
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- #1

- 178

- 0

(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.

- #2

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- #3

- 178

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- #4

vsage

See next post.

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- #5

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- #6

vsage

schattenjaeger said:

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.

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- #7

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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.

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.

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- #8

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As for the guass 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

- #9

Hurkyl

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For small values of y, sin y = y, cos y = 1, and tan y = y. :tongue2:

- #10

vsage

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

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

- #11

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

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

- #12

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Oh right

ok, thanks!

ok, thanks!

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