Solving Limits of (sinx+siny)/(x+y) as (x,y) Approaches (0,0) and (π/3,-π/3)

[Dvsdvs]

f(x,y)=(sinx+siny)/(x+Y)
as (x,y) approaches (0,0) and then for part II (pi/3,-pi/3)

I know that sin(x+Y)/(x+y) would=1 by some simple tweaks. But in my problem, the 2 sins on the numerator are confusing me a little. Since x and y are approaching the same point on the first limit can i say x=y. and write f(x,y)=(sinx+sinx)/2x ?

 

[Dick]

There is a trig identity that will let you express sin(x)+sin(y) in terms of sin((x+y)/2) and cos((x-y)/2), can you dig it up? Assuming x=y isn't going work for part II) anyway.

 

[Shooting Star]

f(x,y)=(sinx+siny)/(x+Y)
as (x,y) approaches (0,0) and then for part II (pi/3,-pi/3)

I know that sin(x+Y)/(x+y) would=1 by some simple tweaks. But in my problem, the 2 sins on the numerator are confusing me a little. Since x and y are approaching the same point on the first limit can i say x=y. and write f(x,y)=(sinx+sinx)/2x ?

I think the teacher (or the book) wants you to use a formula for sinx+siny, which will give the answer directly.

EDIT: Ooh, close finish with Dick...

 

[Dvsdvs]

lol yeah forgot to check formulas. its sinX + sinY = 2sin[ (X + Y) / 2 ] cos[ (X - Y) / 2 ] and it will prob work with both parts ill check it tomorrow. Also, f(x,y) has a singularity on x+y. questions wants proof that it is/ it is not removable.
Is it true that it is removable by plugging in a value at (0,0) i.e. finding a f(x,y) value that equals the lim f(x,y) as x and y approach 0.
Is this rigorous enough? Thank you for the help!

 

[Dick]

Yes, a singularity is removable if there is a well defined limit as you approach the singularity.