Limit with 0^0 form but multivariable version

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[molgera]

"Along Y=-X" seems to be the bane of pretty every multivariable limit I've ever seen that ends up DNE

[kornelviktor]

Is there any calc 3 limit that exist cuz I havent seen one yet

[L_Ratio_]

Isn't that what we can do in polar coordinates too? Along y=0 means theta=0° so (sin(theta)+cos(theta)=1), and along y=-x is equivalent to putting theta= (90+45)° so (cos(theta)+sin(theta) = 0)

[branwave]

The homework limit can be evaluated with the polar coordinates method resulting in an answer of 1.

[miraj]

You didn't need to bother undoing the polar coordinate work at 3:00. A linear combination of sine and cosine can be rewritten as a scaled and shifted sine function (also a scaled and shifted cosine function since they're just shifts of each other).

In particular, cos(θ) + sin(θ) = [cos(θ)/sqrt(2) + sin(θ)/sqrt(2)]*sqrt(2) = [sin(pi/4)*cos(θ) + cos(pi/4)*sin(θ)]*sqrt(2) = sqrt(2)*sin(θ+pi/4)
From this, you can see that cos(θ) + sin(θ) will be 0 when θ = 3pi/4 or 7pi/4. Approaching from this angle, which corresponds to the line y = -x line that you used later on, results in a limit of 0. But any other angle would result in something nonzero^0 = 1 for θ in [0, 2pi)

[leelkek]

Quick question: can someone explain to me why we cannot say that |cos theta + sin theta|^r^2 does not exist? If we pick theta = 3*pi/4 we should have sin theta=-cos theta. So the expression should converge to 0 while taking theta = 0 it should converge to 1. Where is the mistake in my argument? Thank you very much

[electricgamer_yt]

This doesn’t seem like calculus basics I’m a lil lost lol (I’m in 8th grade but learning calculus slowly)

[leonardobarrera]

Can I know in any way what ideas do you have?
I am gonna tell you mines

[rb]

lim(x, y)-> (0, 0) (x^2 + y^2)^(x+y)

x = 0
lim y->0 (y^2y) = lim y->0 e^(2ylny) = e^(lim y->0 2lny/y^-1) = e^(lim y->0 2y^-1/-y^-2) = e^(lim y->0 -2y) = e^0 = 1

y = -x
lim x->0 (x^2 + (-x)^2)^0 = lim x->0 1 = 1

I'm guessing it exists and it's 1. I'm not proving it lol