anonymous
  • anonymous
Please help find the limit (2 Variables) lim(x,y) —> (0,0) of tan(x^2 +y^2)arctan(1/(x^2 +y^2))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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IrishBoy123
  • IrishBoy123
have you tried just plugging in x,y = 0?
anonymous
  • anonymous
yes. but you will end up with tan(0)arctan(1/0)
anonymous
  • anonymous
can we use Lhopitals rule inside the arctan function

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IrishBoy123
  • IrishBoy123
what is tan \((\pi /2) \) ??
anonymous
  • anonymous
sin(π/2)/cos(π/2)
anonymous
  • anonymous
also 1/0
IrishBoy123
  • IrishBoy123
|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
anonymous
  • anonymous
what is the value of arctan(1/0) ?
IrishBoy123
  • IrishBoy123
\[(2n+1) \dfrac{\pi}{2}\]
IrishBoy123
  • IrishBoy123
|dw:1443254619111:dw|
anonymous
  • anonymous
so the answer is 0, but how do you obtain the value of arctan(1/0)
IrishBoy123
  • IrishBoy123
|dw:1443254723319:dw| you have \[0 \times (2n+1)\dfrac{\pi}{2} = 0\]
anonymous
  • anonymous
Thanks
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
type: "too" :p
IrishBoy123
  • IrishBoy123
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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