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I can't solve some limits, and can't find resources online on how to solve this.

I can use double limits (replace X and Y at the same time) or sucessive limits, replacing one each variable separately.

If the limit doesn't exist, I have to show why.

Could someone help me out? Thank you very much!

Problems:

$$\lim_{x\to 0,y\to 0} \frac{3x - 2y}{2x + 3y}$$

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

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  • $\begingroup$ Have you made any attempt yourself? $\endgroup$ – copper.hat Sep 9 '13 at 23:19
  • $\begingroup$ @copper.hat Yes, I am teaching myself limits so it is hard. $\endgroup$ – Pacha Sep 9 '13 at 23:23
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If you have access to Mathematica, I find it very useful to look at the graph. Although this problem is very solvable on paper as well.

First problem:

Direct substitution won't work, because it's indeterminate. One method is to let t=x,t=y and find the limit of a single variable, t. This makes the limit (t)/(5t) = 1/5. This is effectively approaching the limit along the line x=y.

To try a different line, let t=-x,t=y (the line y=-x). From this approach, the limit is (-5t)/(t) = -5. Since these values are different, the function approaches a different value from different directions, and the limit does not exist. This method will only work to prove a limit's non-existence. You cannot prove that a limit exists with this methods, only disprove it.

Second Problem:

You can basically use direct substitution, just keep in mind that the sin(anything) must be within [-1,1]. As the values approach zero, the limit's upper and lower bounds also go to zero, so you can effectively use the squeeze/sandwich theorem, and then (0)(-1) <= answer <= (0)(1) and 0 <= answer <= 0

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  • $\begingroup$ The second problem isn't that simple. Sin of infinite isn't defined. $\endgroup$ – Pacha Sep 9 '13 at 23:31
  • $\begingroup$ There is no unanimity of definition in the second problem. The answer depends on the precise definition that is being used in your course. $\endgroup$ – André Nicolas Sep 9 '13 at 23:36
  • $\begingroup$ @AndréNicolas I was unaware. I just know that the limit is infinitely close to zero, so for Pacha's class zero should probably be fine, right? Because the limit goes to (0)*(sin(undefined)) so (0)*(-1) <= answer <= (0)*(1) $\endgroup$ – Platatat Sep 9 '13 at 23:40
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    $\begingroup$ If we will decide that the limit exists, then that limit will be simply $0$. The issue is that evvery deleted neighbourhood of $(0,0)$ contains points at which the function is not defined, such as $(0,\epsilon)$, for arbitrarily small non-zero $\epsilon$. $\endgroup$ – André Nicolas Sep 9 '13 at 23:46
  • $\begingroup$ @Platatat There is no such thing as "the limit is infinitely close to zero". The limit is zero or it isn't (unless you are talking about surreal or hyperreal numbers). $\endgroup$ – Tunococ Sep 9 '13 at 23:51

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