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$$\lim_{(x,y ) \to (0,0)} ye^{\frac{−1}{\sqrt{x^2+y^2}}}$$

I have tried many ways so far and I keep getting the limit to be $0$. So far I have set x equal to zero and then y and I got the limit to be $0$. I tried setting $y=x$ and $y=x^2$ and still got zero. What else can I do?

My idea is that when both x and y go to zero, the fraction $\frac{-1}{0.00\ldots1}$ becomes negative infinity and when I raise $e$ to negative infinity it will go to zero. And zero times zero is zero. Is this correct? If yes what about my explanation?

user
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Need_MathHelp
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    Did you try polar coordinates? – Hans Lundmark May 26 '23 at 10:46
  • I have edited my question I don't know if you saw it tho. But anyways now that you said I tried polar coordinates and I got something like $\lim_{r \to 0} r\sin(\theta)e^{\frac{−1}{r}}= 0 \times e^{\frac{-\ifty} = 0 \times 0 = 0$. Did I do it right? – Need_MathHelp May 26 '23 at 10:58
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    Yes this is fine $y\to 0$ and $e^{\frac{−1}{\sqrt{x^2+y^2}}} \to 0$ therefore

    $$ ye^{\frac{−1}{\sqrt{x^2+y^2}}}\to 0\cdot 0=0$$

     – user May 26 '23 at 10:59
  • Take a look to theorem 101 n.5 here – user May 26 '23 at 12:34

3 Answers3

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You can use the estimate $$ \bigg| y e^{-\frac{1}{\sqrt{x^2 + y^2}}}\bigg| \le \sqrt{x^2 + y^2} \cdot e^{-\frac{1}{\sqrt{x^2 + y^2}}}, $$ switch to polar coordinates and compute the limit $$ \lim_{r \to + 0} re^{-\frac{1}{r}} = 0. $$

Virtuoz
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  • But when you do that kind of estimation: how can you be sure that it won't diverge or that they even will behave in similar ways around the point (0,0)? But anyways for other similar problems I am allowed to do taylor approximation right – Need_MathHelp May 26 '23 at 11:02
  • @Need_MathHelp if you switch to polar coordinates immediately, you get $\sin t \cdot re^{-1/r}$. The crucial part is $re^{-1/r}$ since $\sin t$ can be made anything in $[-1,1]$ by approaching $(0,0)$ from different angles. Since $re^{-1/r}$ converges to $0$ when $r\to 0^+$, you see that $re^{-1/r}$ will be good estimate since $|\sin t| \leq 1$. Compare this to the case when you remove the minus sign from the exponent. I'll leave it as an exercise for you to think about it. This time $\sin t$ will be important as well. Just remember that $\sin t$ can be anything in $[-1,1]$. – Ennar May 26 '23 at 11:13
  • At the end we are using the same reasoning we can use for the original expression without any new insight, that is $0\cdot 0=0$. – user May 26 '23 at 11:41
  • @user, yes but the technique has broader uses. As I wrote in my previous comment, it's enough to remove the minus sign from the exponent to get the more interesting $0\cdot \infty$. – Ennar May 26 '23 at 11:46
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    @Ennar I agree in general or for the variant you are suggesting, in this specific case I really think it is more confusing than useful. Moreover OP is asking for the direct solution $0\cdot 0=0$. Why use more complicated methods or manipulation? – user May 26 '23 at 11:55
  • @Ennar okey so that is what Virtuz did. As in they just factored out the sin(t) since it could be seen as constant [-1,1]. – Need_MathHelp May 26 '23 at 13:32
  • @Need_MathHelp the point I wanted to make is that the essential part is that $re^{-1/r}$ converges to $0$. If it converged to anything else, the limit wouldn't exist because you can look at two different sequences along different lines such that $\sin t = 1$ or $\sin t = 0$ and you'd get different results. – Ennar May 26 '23 at 16:40
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You idea is perfectly fine and I think that hints or suggestions given are a little bit confusing. Indeed squezee theorem and polar coordinates are really not necessary in this case.

We can simply observe that that both terms goes to zero $y\to 0$ and $e^{\frac{−1}{\sqrt{x^2+y^2}}} \to 0$ to conclude that also their product goes to zero

$$ye^{\frac{−1}{\sqrt{x^2+y^2}}} \to 0\cdot 0=0$$

this property holds in general and follows directly from the definition of limit, there is no reason to use other methods.

The general property we are using is the following, let $f,g: \mathbb R^n \to A\subseteq \mathbb R$ with

$$\lim_{P\to P_0} f(P)=l\in\mathbb R \;\land\;\lim_{P\to P_0} g(P)=m\in\mathbb R \implies \lim_{P\to P_0} f(P)g(P)=lm$$

which directly follows from the definition of limit.

user
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    You are right, but in general it's good to think of polar coordinates whenever you see $\sqrt{x^2+y^2}$ as it can simplify the problem to a limit in single variable which is quite useful in more complicated cases than this one. – Ennar May 26 '23 at 11:43
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    I agree in this case is really not necessary and doesn’t add nothing new for a direct resolution on the original expression. – user May 26 '23 at 11:49
  • If you don't know that $e^{−1/\sqrt{x^2+y^2}} \to 0$ then you would need to prove it. But it is fine to use that fact if you know it (or prove it), and it simplifies the rest of the problem. (The extra factor of $y$ doesn't change the result.) – David K May 26 '23 at 16:00
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Consider polar coordinates: $$\begin{cases} x=\rho \cos \vartheta\\ y=\rho \sin \vartheta \end{cases} $$ If $(x,y)\to(0,0)$, then $\rho \to 0^+$: $$|ye^{-\frac{1}{\sqrt{x^2+y^2}}}|=|\rho \sin \vartheta e^{-1/\rho}|\le|\rho e^{-1/\rho}| \to 0 \quad \text{as} \quad \rho \to 0^+ $$

Sine of the Time
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