Prove that $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2+5xy}{x-y}$ does not exist $$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2+5xy}{x-y}$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic. 
Hints? Thank you.
 A: The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.
More precisely: Each ball of radius $\delta$ around $(0,0)$ contains points with arbitrarily large function values. Consider a sequence of points within the ball that converges towards $(\frac12\delta,\frac12\delta)$ from the $x>y$ side. At points in this sequence the numerator of the fraction goes towards $\frac74\delta^2$, while the denominator goes to $0$ from above, so the function values become abitrarily large.
(Conversely, considering a sequence that approaches $(\frac12\delta,\frac12\delta)$ from the $x<y$ side, will give you arbitrarily large negative function values).
A: Write 
\begin{align}
\dfrac{x^2+y^2+5xy}{x-y} &= x-y +\dfrac{7xy}{x-y} \\
&=x-y+y\left(\dfrac{7x-7y+7y}{x-y}\right) \\
&=x-y+7y+\dfrac{7y^2}{x-y}=x+6y+\dfrac{7y^2}{x-y}.
\end{align}
To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.
A: Although there is an already accepted answer, here is a more systematic method:


*

*Set $y = x-h(x)$ where

*$h$ is continuous and $\lim_{x\to 0} h(x) = 0$
This reveals
$$\frac{x^2+y^2+5xy}{x-y} =\frac{x^2 + (x-h(x))^2 + 5x(x-h(x))}{h(x)}= \frac{7x^2 - 7xh(x) + (h(x))^2}{h(x)}$$
$$ = \color{blue}{\frac{7x^2}{h(x)}} - 7x + h(x)$$
The $\color{blue}{\mbox{blue}}$ term reveals of what order $h(x)$ should converge to $0$ to produce a path $(x,y(x)) \stackrel{x \to 0}{\longrightarrow}(0,0)$ such that the limit for $x \rightarrow 0$ does not exist.
For example $h(x) =x^4 \Rightarrow y=x-x^4$ produces a limit of $+\infty$.
A: A coordinate transformation can work - but you just have to get a little bit more ingenious. The other answers have shown that the trick is the line $x = y$, so we should choose a transformation which makes analyzing it a bit more palatable. Consider the transformation to a tilted coordinate system in which the $y$-axis lies along this line. There are many ways to do this, one of which is the transformation matrix
$$\mathbf{T} := \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix}$$
Geometrically, this matrix is the rotation of the coordinate system around the origin counterclockwise by 45 degrees, thus turning the line $x = y$ counterclockwise just the right amount to align with the $y$-axis (as it lies at a 45-degree angle) multiplied by a scaling by $\sqrt{2}$ to make the entries integer and so easy to work with. Under this transformation, a coordinate vector $(x, y)$ transforms as
$$(x, y) \mapsto (x - y, x + y)$$
That is, define $u := x - y$ and $v := x + y$ as your new coordinates. Then the denominator becomes $u$, and the numerator is solved as follows: note that $v^2 = (x^2 + 2xy + y^2)$ so
$$x^2 + y^2 + 5xy = (x^2 + 2xy + y^2) + 3xy = v^2 + 3xy$$
and now, by recognizing the sum-and-difference-of-the-same-thing pattern, that you can take $u + v = 2x$ and $u - v = -2y$ so $(u + v)(u - v) = -4xy$ and $\frac{3}{4} (u + v)(u - v) = 3xy$. Thus
$$\begin{align}\frac{x^2 + y^2 + 5xy}{x - y} &= \frac{v^2 - \frac{3}{4}(u + v)(u - v)}{u} \\
&= \frac{v^2 - \frac{3}{4}(u^2 - v^2)}{u} \\
&= \frac{v^2 - \frac{3}{4}u^2 + \frac{3}{4}v^2}{u}\\
&= \frac{-\frac{3}{4}u^2 + \frac{7}{4}v^2}{u}\\
&= -\frac{3}{4} u + \frac{7}{4} \frac{v^2}{u} \\
&= \frac{7}{4} \frac{v^2}{u} - \frac{3}{4} u
\end{align}$$
and now when you take the limit - note that the origin is unchanged, so the limit is still to $(0, 0)$ but now in $(u, v)$ coordinates instead of $(x, y)$ coordinates - the second term is zero, but the first has no limit: if $v$ goes to 0 first, then the limit as $u$ goes to 0 is 0, but if $u$ goes to 0 first, the limit becomes $\pm \infty$, and is such for every value of $v$ except 0 for such an approach. That is, the function is unboundedly large in an arbitrarily small circle around the origin, and thus the limit at that origin does not exist.
