Limit with 0^0 form but multivariable version

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Күн бұрын

I asked this limit question on my Instagram: www.instagram.... and only 46% got this right. Can we use polar coordinates to help use evaluate this limit or how do we show this limit does not exist.
How to evaluate the limit of a multi-variable function (6 methods you need to know): • 6 methods of evaluatin...
How to show the limit of a multivariable function does not exist: • How to show the limit ...
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Пікірлер: 17

@bprpcalculusbasics Күн бұрын

How to evaluate the limit of a multi-variable function (6 methods you need to know): kzbin.info/www/bejne/fHuQoJmGi75ohposi=FHacMMc3Hi9n3N-C How to show the limit of a multivariable function does not exist: kzbin.info/www/bejne/r3izpKKcicmnh9k

@miraj2264 Күн бұрын

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)

@bra1nwave172 Күн бұрын

Can you explain how to do this for the nontrivial case: asin(θ) + bcos(θ)?

@miraj2264 Күн бұрын

@@bra1nwave172 It's a bit messy to write in text, but hopefully this is readable: asin(θ) + bcos(θ) = sqrt(a^2 + b^2)*{[a/sqrt(a^2 + b^2)]*sin(θ) + [b/sqrt(a^2 + b^2)]*cos(θ)} Now draw a right triangle, with legs of length a and b. Then the hypotenuse has length sqrt(a^2 + b^2). Call the angle between the leg of length a and the hypotenuse x. Then cos(x) = a/sqrt(a^2 + b^2) and sin(x) = b/sqrt(a^2 + b^2). Substituting back in to the equation above, you get: sqrt(a^2 + b^2)*(cos(x)sin(θ) + sin(x)*cos(θ)) = sqrt(a^2 + b^2)*sin(θ+x)

@bra1nwave172 Күн бұрын

@@miraj2264 Beautiful result. I wanted to know this so I have an alternative method to solving integrals of rational functions of sine and cosine when the substitution isn't u = sinx,cosx,tanx.