Limit of $\frac{\cos{x}-1-\frac{x^2}{2}}{x^4+y^4}$ as it goes to the origin How can I solve a limit like this one:
$$\lim_{(x,y) \rightarrow (0,0)}\frac{\cos{x}-1-\frac{x^2}{2}}{x^4+y^4}$$
If I do $x=0$ so it aproaches through $y$ or $y=0$ so it approaches through $x$ it doesn't really help because of the division over $0$ in the denominator. As well as $x=y$, it doesn't make things better. 
Thanks for the help. 

If $x = 0$ there is no division by zero ! Since when $x = 0$ 
$$f(x ,y) = \frac{\cos(x) - 1 -x^2/2}{x^4 + y^4} = \frac{1-1 - 0}{y^4} = 0$$
Therefore 
$$ \lim_{y \to 0} f(0,y) = \lim_{y \to 0} 0 = 0.$$
If $y = 0$ then 
$$ f(x,0) = \frac{\cos(x)-1-x^2/2}{x^4}.$$
Since the numerator and the denominator both converge to $0$ we can use L'Hospital's rule :
$$ \frac{- \sin x - x}{4x^3}$$
Since the numerator and denominator both converge to $0$ we use L'Hospital's rule a second time
$$ \frac{- \cos x - 1}{12x^2}$$
so we have 
$$ \lim_{x \to 0} f(x,0) = \lim_{x \to 0} \frac{-\cos(x)-1}{12x^2} = - \infty$$
Since the limit of $(x,y)$ to $(0,0)$ is different along two different paths the limit
$$\lim_{(x,y) \to (0,0)} \frac{\cos(x)-1-x^2/2}{x^4 + y^4}$$
does not exist !
