# If I show that a multivariable limit goes to $\infty$, then is this enough to show that the limit does not exist?

For example, consider the limit:

$$\lim_{(x,y)\to(0,0)}\dfrac{1}{x^2+y^2}$$

Plugging in $x=0, y=0$ will give us the form $1/0$ which is infinity.

Would this be enough to show the limit does not exist or do I have to consider each side?

Well, in this particular case, as $\lVert (x,y)\rVert\to 0$, we have that the limit goes to infinity. This isn't too hard to see, because of the radial symmetry of $f(x,y)=x^2+y^2$. In general, however, this depends on what you mean by "exists." The limit does not exist in the sense that it is not a real number.

• Well suppose I said evaluate the limit. WHat would you say? Would you say it exists or doesnt – K Split X Oct 14 '17 at 22:14
• You can't evaluate the limit, in some sense. – Alekos Robotis Oct 14 '17 at 22:14
• Thats my question, so would you automatically say it does not exist? – K Split X Oct 14 '17 at 22:14
• I would say that the limit does not exist. In particular, the limit is not a real number. – Alekos Robotis Oct 14 '17 at 22:15
• "we have that the limit goes to infinity" Limits don't go anywhere. They either exist or don't. – zhw. Oct 14 '17 at 22:31

If we think about the analogous one variable limit $\lim_{x\to 0}\left(\dfrac{1}{x^2}\right)=+\infty,$ this value exist as an extended real number, but not as a real number. Same thing happens for this function. To show $\lim_{(x,y)\to(0,0)}\dfrac{1}{x^2+y^2}=+\infty,$ take any sequence of points $(x_n, y_n)_{n\in\Bbb{N}}$ that converge to $(0,0)$ and use the inequality $\dfrac{1}{2\max(x_n^2,y_n^2)}\le\dfrac{1}{x_n^2+y_nn^2}.$