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I’m having a difficult time trying to find two different paths that give me different limits for the following:

$$\lim_{(x,y) \rightarrow (0,0)} \frac{x-y}{x^2 + y^2}$$

- Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$
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- limit $\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$

- Finding the Radius of a Circle in 3D Using Stokes Theorem
- Show $\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$
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- Convergence of a recursively defined sequence
- Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$
- How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$?

For reasons that will become apparent further down, I find it useful to instead study $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{x^2+y^2}{x-y}\right).$$

Note that for all $(x,y)\in \mathbb R^2$ with $x\neq y$ it holds that $$\dfrac{x^2+y^2}{x-y}=\dfrac{2x^2-(x^2-y^2)}{x-y}=\dfrac{2x^2}{x-y}-(x+y).$$

Since $\lim \limits_{(x,y)\to (0,0)}(x+y)$ exists, any problems that might occur with the initial limit will occur on $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2}{x-y}\right).$$

This last limit is fairly simple to manipulate by choosing $y$ as a function of $x$. If you stare at the fraction long enough, and hopefully not for too long, you’ll see that setting $y=x-kx^2$ with $k\in \mathbb R$ will yield a $k$-dependent limit.

You can now abandon this scratch work and consider the paths $t\mapsto (t,t-kt^2)$ in the initial limit.

For some problems with such a great degree symmetry in $x$ and $y$ you can try working in polar coordinates

\begin{equation*}

\begin{cases}

x = r \cos(\theta) \\

y = r \sin(\theta)

\end{cases}

\end{equation*}

The statement $(x,y) \rightarrow (0,0)$ is equivalent to $r \rightarrow 0$, and by taking $\theta = \theta(r)$ you get any path you want.

Now, the limit in terms of polar coordinates is

\begin{equation*}

\lim_{r \rightarrow 0}\frac{\cos(\theta) – \sin(\theta)}{r}

\end{equation*}

If you pick a path with $\theta$ such that $\cos(\theta)\neq \sin(\theta)$, you get $\infty$.

If you pick a path with $\theta$ such that $\cos(\theta) = \sin(\theta)$, you get $0$.

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