Consider the function...

Question

1. Please help. I cannot upload a photo, so I have tried to translate the limit of F(x,y) just using my keyboard.

Question:

Consider the function 𝑓(𝑥,𝑦)=(𝑥^𝑚)(𝑦^3)/((𝑥^2)+(𝑦^4)), where 𝑚∈𝐙 is a fixed integer parameter (we only take integer values to avoid the problem of definability when 𝑥 < 0). For which values of 𝑚∈𝐙 does the limit along any path 𝑦 = 𝑘𝑥 is equal to 0?

In other words, find all 𝑚∈𝐙 for which we have that

for all 𝑘∈𝐑, ((lim(𝑥,𝑦)→(0,0)) 𝑦=𝑘𝑥) ((𝑓(𝑥,𝑦)=lim𝑓(𝑥,𝑘𝑥)=0.) 𝑥→0)

1.

For 𝑚>1

2.

For 𝑚>−1

3.

For 𝑚∈{−1,0,1}

4.

For 𝑚<1

5.

For 𝑚<−1

6.

For no 𝑚∈𝐙 is the limit at (0,0) along every path 𝑦=𝑘𝑥 equal to 0.

Answer 1

To start, let's sub in the y=kx to get f(x,y) only in terms of x. This makes the equation k³x^3+m / (x² + k⁴x⁴). Now depending on what m is, we can simply use l'Hospital's Rule to find the limit. Bear in mind that we can only use l'Hospital's rule a maximum of 2 times before the denominator becomes a constant (2 + 12k⁴x² for x=0). I'm assuming that based on your parsing that you're looking for which of the 5 options gives you no limit = 0 (let me know if this is wrong and I can edit my response. So let's look at values around the cutoff points.

m=1: For this we get 12k³x² when that constant is reached, so this gives us a limit of zero.

m=-1: For this we get a numerator of 2k³ when that constant is reached. This gives us a limit of k³ which can be zero if k=0.

m<-1: In the special case that m∈{-2, -3}, we get a constant in the numerator before the denominator becomes a constant. If k≠0, this limit will approach ±∞. If k=0, we can simply take the limit as k→0 and note that the numerator becomes a constant before the denominator, making this c/0 going towards ±∞.

In the case where m<-3, we get a value approaching ±∞ in the numerator and 0 in the denominator. This approaches ±∞ as well.

Thus, the only solution that gives you a non-zero limit is m<-1.