How to check if the limit exists in multivariable calculus

Question

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

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?

Answer 1

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.

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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.

Answer 2

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 $$ \lim_{r \to + 0} re^{-\frac{1}{r}} = 0. $$

Answer 3

Consider polar coordinates: $$\begin{cases} x=\rho \cos \vartheta\\ y=\rho \sin \vartheta \end{cases} $$ If $(x,y)\to(0,0)$, then $\rho \to 0^+$: $$|ye^{-\frac{1}{\sqrt{x^2+y^2}}}|=|\rho \sin \vartheta e^{-1/\rho}|\le|\rho e^{-1/\rho}| \to 0 \quad \text{as} \quad \rho \to 0^+ $$