A couple months ago I had a math test which I couldn't do this two-part exercise,
Given $f(x,y)=\frac{(x-1)^2(y-1)}{(x-1)^4+(y-1)^2}$ and $g(x,y)=\frac{(x-1)^2(y-1)^2}{(x-1)^4+(y-1)^2}$
So the question for both parts was find, if it exists, the limit as $(x,y)\rightarrow(1,1)$
Today after reading a little about methods for calculating limits I thought of the following solution:
1) $0\le\lvert f(x,y)\rvert = \frac{(x-1)^2\lvert(y-1)\rvert}{(x-1)^4+(y-1)^2}$
2) $(x-1^2)\le (x-1)^4 + (y-1)^2 \rightarrow \frac{(x-1)^2}{(x-1)^4+(y-1)^2} \le 1 \rightarrow \frac{(x-1)^2}{(x-1)^4+(y-1)^2}*\lvert(y-1)\rvert \le \lvert (y-1)\lvert$
Therefor $\lim \limits_{x,y \to 1,1}0\le \lim \limits_{x,y \to 1,1}\frac{(x-1)^2\lvert(y-1)\rvert}{(x-1)^4+(y-1)^2} \le \lim \limits_{x,y \to 1,1}\lvert (y-1)\lvert$ $\rightarrow 0 \le \lim \limits_{x,y \to 1,1}\frac{(x-1)^2\lvert(y-1)\rvert}{(x-1)^4+(y-1)^2} \le 0$
So $\lim \limits_{x,y \to 1,1}|f(x,y)| = 0 \rightarrow \lim \limits_{x,y \to 1,1}f(x,y) = 0$
And then again the exact same proof for $g(x,y)$ since the only difference is on $(y-1)$ being squared on the second function. So my question is, is this correct? I mean, it seems weird to me to have two exercises in a single exam with exact same solution
$\endgroup$2 Answers
$\begingroup$It is easier to change the function to $h(x,y) = f(x+1, y+1)$ and limit to $(x,y) \to (0,0)$ and then use polar coordinates:
$$\lim_{(x,y)\to (0,0)} {x^2y\over x^4+y^2} = \lim_{r \to 0} {r^3 \cos^2(\theta)\sin(\theta)\over r^2(r^2 \cos^4(\theta) + \sin^2(\theta))} = \lim_{r \to 0} {r \cos^2(\theta)\sin(\theta)\over r^2 \cos^4(\theta) + \sin^2(\theta)} = 0$$
Since $\lim_{(x,y)\to (0,0)}h(x,y) = 0$, also $\lim_{(x,y)\to (1,1)}f(x,y) = 0$
Analogously proceed for $g(x,y)$.
$\endgroup$ 1 $\begingroup$Neither method presented here is actually correct. The limit of the first function does not exist. I will also shift my variables and look at $$ h(x,y) = \frac{x^2y}{x^4 + y^2}. $$ First, let's look at $y = x$, we have $$ \lim_{x \to 0} h(x,x) = \lim_{x \to 0} \frac{x^3}{x^4 + x^2} = \lim_{x \to 0} \frac{x}{1+ x^2} = 0 .$$ On the other hand, look at $x^2 = y$. Then, $$ \lim_{y \to 0} h(x, x^2) = \lim_{x \to 0} \frac{x^4}{2x^4} = \frac{1}{2}. $$ Since these do not agree, the limit does not exist. It might be illuminating for you to plot the graph of this function.
For the second one, you can show that the limit is zero from the squeeze theorem.
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