## 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))

1. IrishBoy123

have you tried just plugging in x,y = 0?

2. anonymous

yes. but you will end up with tan(0)arctan(1/0)

3. anonymous

can we use Lhopitals rule inside the arctan function

4. IrishBoy123

what is tan $$(\pi /2)$$ ??

5. anonymous

sin(π/2)/cos(π/2)

6. anonymous

also 1/0

7. 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

8. anonymous

what is the value of arctan(1/0) ?

9. IrishBoy123

$(2n+1) \dfrac{\pi}{2}$

10. IrishBoy123

|dw:1443254619111:dw|

11. anonymous

so the answer is 0, but how do you obtain the value of arctan(1/0)

12. IrishBoy123

|dw:1443254723319:dw| you have $0 \times (2n+1)\dfrac{\pi}{2} = 0$

13. anonymous

Thanks

14. IrishBoy123
15. IrishBoy123

type: "too" :p

16. 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 $$w = \dfrac{1}{z}$$ $$\implies lim_{z \to 0} \tan z \times \,lim_{w \to +\infty} \arctan w$$ $$lim_{z \to 0} \tan z = 0$$ $$\lim_{w\to\frac\pi 2}\tan w=+\infty\iff \lim_{w\to+\infty}\arctan w=\frac\pi2$$ $$\therefore lim_{x,y \to 0,0} \, \tan(x^2 +y^2).\arctan(\dfrac{1}{x^2 +y^2}) = 0$$