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$\lim\limits_{(x,y) \to (0,0)} f(x,y) = \dfrac{\cos(x) -1 - {x^2 \over 2}}{x^4 + y^4}$

Is the following approach correct?

If we approach the origin from $y$ , that is $x = 0$:

$\lim\limits_{(x=0,y) \to (0,0)} f(0,y) = {0 \over y^4} = 0$

Now we approach the origin from $x$ and use $\cos(x) \sim_{0} 1 - {x^2\over 2}$

$\lim\limits_{(x,y=0) \to (0,0)} f(0,y) = { -x^2 \over x^4} =- {1 \over x^2}=-\infty$

Then we can conclude that the limit does not exist.

My teacher took another approach where he uses the $\cos (2x)$ formula and ends up using $\lim\limits_{u\to0}\dfrac{\sin u}{u}=1$ so I'm wondering if my solution is OK or not.

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    $\begingroup$ Is the numerator, by any chance, intended to contain $\cos(x) - (1-\frac {x^2}2)$ in place of what you wrote? $\endgroup$ – Lord_Farin May 28 '13 at 11:59
  • $\begingroup$ @Lord_Farin I doubled checked the problem statement from the text book and it is as I posted it. $\endgroup$ – jjjx May 28 '13 at 12:04
  • $\begingroup$ have you tried the l'Hopital rule? $\endgroup$ – user108205 Dec 7 '13 at 16:25
  • $\begingroup$ See also math.stackexchange.com/questions/418208/… $\endgroup$ – Martin Sleziak Sep 2 '14 at 17:42
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Use the straight line $y=mx$ for $m \in \mathbb{R}$: $$ \lim_{x \to 0} \frac{\cos x -1 -\frac{x^2}{2}}{(1+m^2)x^2} = \lim_{x \to 0} \frac{1-\frac{x^2}{2}+\frac{c^4}{4!}-1-\frac{x^2}{2}}{(1+m^2)x^2}=\lim_{x \to 0} \frac{-x^2+\frac{c^4}{4!}}{(1+m^2)x^2} $$ for some $c \in (0,x)$. Therefore $$ \lim_{x \to 0} \frac{-x^2+\frac{c^4}{4!}}{(1+m^2)x^2} = -\frac{1}{1+m^2}, $$ which clearly depends on $m$. The original limit cannot exist.

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My hint:

Apply Taylor expand, we have: $\cos x-1-\frac{x^2}{2}=-x^2+O(x^2)$

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

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  • $\begingroup$ What is $O(x^2)$? $\endgroup$ – user108205 Dec 7 '13 at 16:26
  • $\begingroup$ $O(x^2)$ should be $o(x^2)$ and the last $=-\infty$ does not hold. $\endgroup$ – Did May 22 '14 at 8:56

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