How to find $\lim_{(x,y) \to (0,0) } \frac{x+2y}{\sqrt{x^2+y^2}}$? How to find:
$$\lim_{(x,y) \to (0,0) }\frac{x+2y}{\sqrt{x^2+y^2}}\,\,?$$
I tried to use polar transformation but the $r$'s cancel out
 A: The fact that the $r$’s cancel actually makes it easy: you find that you want
$$\lim_{r\to 0}(\cos\theta+2\sin\theta)\;,$$
and this plainly doesn’t exist, since the limit is $1$ when $\theta=0$ and $2$ when $\theta=\frac{\pi}2$.
Alternatively, it’s often worth trying to approach the origin along paths of the form $y=kx$. Here you get
$$\lim_{x\to 0}\frac{x+2kx}{\sqrt{x^2+(kx)^2}}=\lim_{x\to 0}\frac{(2k+1)x}{x\sqrt{k^2+1}}=\frac{2k+1}{\sqrt{k^2+1}}\;,$$
and it’s easy to check that this depends on $k$.
A: To show that a multivariable limit doesn't exist, you just need to prove that the limit
isn't the same for any two directions. One way to do this is to go from the $x$ direction (set $y = 0$ and find the limit), and then the $y$ direction (set $x = 0$ and find the limit). If these two limits aren't the same, then you are done.
If both limits are the same, then you can try testing $y=x$ or $y=kx$ as shown in Brian's answer. You can also convert to a different coordinate system to see if the problem will be easier in polar coordinates. In your case, all you need to do is test either the $x$ direction or the $y$ direction.
For the $x$ direction we approach $(0,0)$ along the $x$-axis,
$$\displaystyle\lim_{(x,y) \to (0,0) } \frac{x+2y}{\sqrt{x^2+y^2}}=\displaystyle\lim_{x\to 0}\frac{x}{\sqrt{x^2}}=\begin{cases}
 1,&\text{if}\, x\to 0^+,\\
 -1,&\text{if}\ x\to 0^-.\\
\end{cases}$$
For the $y$ direction we approach $(0,0)$ along the $y$-axis,
$$\displaystyle\lim_{(x,y) \to (0,0) } \frac{x+2y}{\sqrt{x^2+y^2}}=\displaystyle\lim_{y\to 0}\frac{2y}{\sqrt{y^2}}=\begin{cases}
 2,&\text{if}\, y\to 0^+,\\
 -2,&\text{if}\ y\to 0^-.\\
\end{cases}$$
Therefore, the limit doesn't exist because in the $x$ direction $\underset{x->0^-}{\lim} f(x)\neq \underset{x->0^+}{\lim} f(x)$ and similarly in the $y$ direction. You could also conclude that the limit doesn't exist because the limit in the $x$ direction is different than the limit in the $y$ direction.
