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anonymous

  • one year ago

Please help find the limit (2 Variables) lim(x,y) —> (0,0) of tan(x^2 +y^2)arctan(1/(x^2 +y^2))

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  1. IrishBoy123
    • one year ago
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    have you tried just plugging in x,y = 0?

  2. anonymous
    • one year ago
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    yes. but you will end up with tan(0)arctan(1/0)

  3. anonymous
    • one year ago
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    can we use Lhopitals rule inside the arctan function

  4. IrishBoy123
    • one year ago
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    what is tan \((\pi /2) \) ??

  5. anonymous
    • one year ago
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    sin(π/2)/cos(π/2)

  6. anonymous
    • one year ago
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    also 1/0

  7. IrishBoy123
    • one year ago
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    |dw:1443254278447:dw| tan pi/2 is undefined and then that behavious repeats itself meaning that arctan(1/0) is defined so you have 0 x something finite = 0

  8. anonymous
    • one year ago
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    what is the value of arctan(1/0) ?

  9. IrishBoy123
    • one year ago
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    \[(2n+1) \dfrac{\pi}{2}\]

  10. IrishBoy123
    • one year ago
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    |dw:1443254619111:dw|

  11. anonymous
    • one year ago
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    so the answer is 0, but how do you obtain the value of arctan(1/0)

  12. IrishBoy123
    • one year ago
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    |dw:1443254723319:dw| you have \[0 \times (2n+1)\dfrac{\pi}{2} = 0\]

  13. anonymous
    • one year ago
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    Thanks

  14. IrishBoy123
    • one year ago
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    if that's to brutal http://math.stackexchange.com/questions/711194/how-to-prove-that-limit-of-arctanx-as-x-tends-to-infinity-is-pi-2

  15. IrishBoy123
    • one year ago
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    type: "too" :p

  16. IrishBoy123
    • one year ago
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    if you want some formalism, this makes sense, but i think it might be OTT: \(lim_{x,y \to 0,0} \, \tan(x^2 +y^2).\arctan(\dfrac{1}{x^2 +y^2})\) let \(z = x^2 + y^2\) \(lim_{x,y \to 0,0} \, z = 0\) \(\implies lim_{z \to 0} \tan z . arctan \dfrac{1}{z}\) \(= lim_{z \to 0} \tan z \times \,lim_{z \to 0} \arctan \dfrac{1}{z}\) let \(w = \dfrac{1}{z}\) \(\implies lim_{z \to 0} \tan z \times \,lim_{w \to +\infty} \arctan w\) \( lim_{z \to 0} \tan z = 0\) \(\lim_{w\to\frac\pi 2}\tan w=+\infty\iff \lim_{w\to+\infty}\arctan w=\frac\pi2\) \(\therefore lim_{x,y \to 0,0} \, \tan(x^2 +y^2).\arctan(\dfrac{1}{x^2 +y^2}) = 0\)

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