# Determine whether a multi-variable limit exists $\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$

I need to determine whether the next limit exists:

$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$$

Looking at the numerator $(-1-\frac{x^2}2)$ it immediately reminds me of maclaurin series of $\cos$, Where:

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...=1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)$$

So we can simplify the expression:

\begin{align} \lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}&=\lim_{(x,y)\to(0,0)}\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)-1-\frac{x^2}2}{x^4+y^4}\\ &=\lim_{(x,y)\to(0,0)}\frac{-x^2+\frac{x^4}{4!}+O(x^6)}{x^4+y^4} \end{align}

This limit does not exist.

So, after solving it the 'hard' way, I was thinking:

If I place $y=mx$ and then try to solve the limit I get:

$$\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4(1+m^4)}=\infty$$

Is that enough to prove that the limit does not exists? I mean, If I find at least one limit that does not 'converge' is it enough to say that a multi-variable function limit doesn't exist?

• If a multi-variable function has a (finite) limit at a point, then the image of every sequence converging to the point doesn't depend on the sequence. Hence if you plug in $y=mx$ and get limit $\infty$, then either the limit doesn't exist, or it is $\infty$. – ronash Jun 12 '13 at 8:36
• Perhaps $\cos x-(1-\frac{x^2}{2})$ was intended. Still, limit does not exist. – André Nicolas Jun 12 '13 at 8:40
• So i need to make sure there's a another 'curve' that has a limit different than $\infty$ to make sure it does no exist? – StationaryTraveller Jun 12 '13 at 8:44
• And anyhow, The first way i did is enough to prove the limit does not exist? – StationaryTraveller Jun 12 '13 at 8:45
• – Martin Sleziak Sep 2 '14 at 17:42

$$\lim_{n\to\infty}f(x,y_n),~~y_n=\pi/n$$
Find the limit as $x \to 0$ when $y=x$ and show that it is not equal the limit as $x\to 0$ when $y=0$.