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Evaluate $\lim_\limits{(x,y)\to(0,0)}\dfrac{\cos(xy)-1}{x^2y^2}$

The limit does exist.

The only thing I can think of is let $t=xy$. Then our limit becomes

$$\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}=\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}\dfrac{(\cos(t)+1)}{(\cos(t)+1)}=\lim_{t\to 0}\dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}$$

This is how far I got and then I don't know. Not even sure if this is the best method to find the limit.

Any hints would help.

Thank you.

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  • $\begingroup$ Use the power series of $\cos(t)$ $\endgroup$ – TomGrubb Sep 25 '17 at 23:49
  • $\begingroup$ Not really familiar with that concept. Any other ideas? $\endgroup$ – K Split X Sep 25 '17 at 23:49
  • $\begingroup$ Are you sure the limit exists? As the question's currently written there are lines along which one can approach (0,0) on which the function is undefined. $\endgroup$ – Steven Stadnicki Sep 25 '17 at 23:53
  • $\begingroup$ According to Wolfram alpha the limit is $-1/2$ $\endgroup$ – K Split X Sep 25 '17 at 23:53
  • $\begingroup$ You Can't Trust Alpha. Especially in matters of definability; it will cheerfully give you an answer it thinks is correct, without really confirming whether all the operations make semantic sense. $\endgroup$ – Steven Stadnicki Sep 25 '17 at 23:57
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You are basically done,

I would break the expression inside the limit into,

$$\frac{-1}{1+\cos t} \frac{\sin t}{t} \frac{\sin t}{t}$$

Then use the product rule for limits and the famous limit $\frac{\sin t}{t} \to 1$ as $t \to 0$.


Edit

As per the answer of @Steven Stadnicki, your limit is only equal to the single variable limit if $xy \neq 0$ as we approach the origin. Otherwise, the expression is not even defined even if we are very close to the origin. It's safer to say that $f(x,y)=\frac{\cos (xy)-1}{(xy)^2}$ tends to negative a half as we approach the origin going through points which are a part of the domain of $f$ in which the function is defined.

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  • $\begingroup$ I think this is the easiest to understand. Thanks $\endgroup$ – K Split X Sep 25 '17 at 23:55
  • $\begingroup$ Np :)${}{}{}{}{}$ $\endgroup$ – Ahmed S. Attaalla Sep 26 '17 at 0:29
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You know that $$\lim_{t\to 0} \frac{\cos t-1}{t^2} =-\frac{1}{2}, $$using L'Hospital, for example. This tells us that $g\colon \Bbb R \to \Bbb R$ defined by $$g(t) \doteq \begin{cases} \dfrac{\cos t-1}{t^2}, \mbox{if } t \neq 0 \\[.5em] -\dfrac{1}{2}, \mbox{if }t=0\end{cases}$$is continuous. Because of that, we can write $$\lim_{(x,y)\to (0,0)} \frac{\cos (xy)-1}{x^2y^2} = \lim_{(x,y) \to (0,0)} g(xy) = g\left(\lim_{(x,y)\to (0,0)}xy\right) = g(0) = -\frac{1}{2}.$$

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Hint. One may recall that, as $t \to 0$, $$ \cos t \to 1,\qquad \frac{\sin t}{t} \to 1, $$ giving, as $t \to 0$, $$ \dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}\to -\frac12. $$

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  • $\begingroup$ Not really a hint.. $\endgroup$ – Ahmed S. Attaalla Sep 25 '17 at 23:55
  • $\begingroup$ @AhmedS.Attaalla essentially its the same answer as yours, using the product rule $\endgroup$ – K Split X Sep 25 '17 at 23:56
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To expand my comments above into a somewhat contrarian answer: I would argue that the limit above, as written, does not exist. This is because the definition of limit requires that the function be defined on a punctured neighborhood of the point in question; more specific to this case, we would say that the limit as $\langle x,y\rangle\to\langle0,0\rangle$ is $L$ iff for every $\epsilon$ there exists a $\delta$ such that for all $\langle x,y\rangle\neq\langle 0,0\rangle$ with $\left|\langle x,y\rangle-\langle0,0\rangle\right|\lt\delta$, $\left|f(x,y)-L\right|\lt\epsilon$. This fails to hold here because there are many $\langle x,y\rangle$ for which $f(x,y)$ (defined as $f(x,y) = \dfrac{\cos(xy)-1}{x^2y^2}$) is undefined — namely, all points of the form $\langle x,0\rangle$ or $\langle 0,y\rangle$.

Note that it's possible to define $f(x,y)$ at points on the axes in such a way that the function is continuous on all of $\mathbb{R}^2-\langle0,0\rangle$: just use the same trick that's given in the answers and say that $f(x,0)=-\frac12$ for $x\neq 0$, and likewise $f(0,y)=-\frac12$ for $y\neq0$. Then the limit of that function as $\langle x,y\rangle\to\langle0,0\rangle$ exists and can be found using the techniques presented in the other answers (with some careful justification). But this isn't how the initial function $f()$ is defined.

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  • $\begingroup$ Good point, I have added this into my answer and credited you if you don't mind. $\endgroup$ – Ahmed S. Attaalla Sep 26 '17 at 0:40
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    $\begingroup$ Nevertheless, the point $(0,0)$ is a limit point of the domain $\{(x,y) \in \Bbb R^2 \mid xy \neq 0\}$, so I really don't see the problem. You take the limit along the set where $f$ is defined, so it should be "for all $(x,y) \in {\rm dom}(f)\setminus \{(0,0)\}$ such that $\|(x,y)-(0,0)\| \leq \delta$..." $\endgroup$ – Ivo Terek Sep 26 '17 at 1:00
  • $\begingroup$ Stephen. Very good point.The original problem is 2D. Setting t_nm=x_ny_m, or t=xy, formally solves the problem. How would you one handle t_mn if you consider sequences?Thank,Peter $\endgroup$ – Peter Szilas Sep 26 '17 at 6:24
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For $xy\ne 0$ we have $$\frac {-1+\cos (xy)}{x^2y^2}=\frac {-2\sin^2(xy/2)}{x^2y^2} =\left(-\frac {1}{2}\right) \left(\frac {\sin (xy/2)}{(xy/2)}\right)^2 .$$ As other answerers have noted, the limit only exists if we restrict both $x$ and $y$ to non-zero values.

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$$\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}=\lim_{t\to 0}\dfrac{\cos(t)-cos(0)}{t^2} $$

$$\lim_{t\to 0}\dfrac{\cos(t)-cos(0)}{t^2}=\lim_{t\to 0}\dfrac{-2\sin(\frac t 2)\sin(\frac t 2 )}{t^2} =\lim_{t\to 0}\dfrac{-2\sin^2(\frac t 2)}{t^2}$$

Since $\lim_{x\to 0}\dfrac{ \sin(x)} {x}=1$ ,

$$\lim_{t\to 0}\dfrac{-2\sin^2(\frac t 2)}{t^2}=\frac {-1} 2\lim_{t\to 0}\dfrac{ \sin^2(\frac t 2)}{\frac {t^2}{4}}=\frac {-1}{2}$$

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