So let's say I wanted to find $$ \displaystyle \lim_{(x,y) \to (0,0)}\dfrac{xy^2} {x^2 +y^4} $$ Let's call the function above $f(x,y)$ then $h(x,y) = 0$ would be a function smaller than $f$ that would work for the sandwich or squeeze rule. What's an easy way to find a function larger than $f(x,y)$ in order to complete the sandwich rule? Is there some specific method? Thanks in advance.
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2The function does not have a limit as $(x,y)\to (0,0)$. – André Nicolas Mar 19 '16 at 06:19
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The limit doesn't exist? That's strange, wolfram alpha says the limit is 0 – Ryan J Mar 19 '16 at 06:39
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The linked file does not seem to compute the limit. I have written an answer in which it is shown the limit does not exist. – André Nicolas Mar 19 '16 at 06:47
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limit (x*y^2) / (x^2 + y^4) as (x,y)->(0,0) That's what I had on wolfram alpha. You're answer does look correct though. I'm a little confused right now – Ryan J Mar 19 '16 at 06:57
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Hint: $$ \begin{align} \frac{xy^2}{x^2 +y^4} &=\frac1{\frac{x}{y^2}+\frac{y^2}{x}}\\ &=\frac{\operatorname{sign}(x)}{\frac{|x|}{y^2}+\frac{y^2}{|x|}}\\ &=\frac{\operatorname{sign}(x)}{\left(\frac{\sqrt{|x|}}{y}-\frac{y}{\sqrt{|x|}}\right)^2+2}\\ \end{align} $$ Consider the paths $(t^2,t)$ and $(-t^2,t)$ as $t\to0$.
robjohn
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Interesting method. BTW in older books you see sgn (x) (Latin : signum) for sign(x). – DanielWainfleet Mar 19 '16 at 07:04
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@RyanAndo: Along the path $(t^2,t)$, the function equals $\frac12$. Along the path $(-t^2,t)$, the function equals $-\frac12$. – robjohn Mar 19 '16 at 08:52
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The function $f(x,y)$ does not have a limit as $(x,y)\to (0,0)$.
To see this, note that if we approach $(0,0)$ along the path $x=y$, the limit is $0$, while if we approach $(0,0)$ along the path $x=y^2$, the limit is $\frac{1}{2}$.
In particular, if we use lower piece of bread $h(x,y)=0$, there is no $k(x,y)$ which approaches $0$ such that $h(x,y)\le f(x,y)\le k(x,y)$.
André Nicolas
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Last question, what gave you the impression that the limit does not exist? Was it after you tried along the path x = y^2. Or were you able to recognize it immediately? If so, how? – Ryan J Mar 19 '16 at 07:50
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Basically from $x=y^2$ and relatives, like $x=10y^2$. But with these things it is sometimes hard to separate out figuring out from remembering. I have taught the material often! – André Nicolas Mar 19 '16 at 07:54
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One of the classical method that you can show that this kind of limits dont exist is this, you use $x=my^k$ or $y=mx^k$ where $m$ is variable and $k$ is integer. After substitution $x=my^k$ or $y=mx^k$ in the limit, if the limit be dependent of the variable $m$, it means, it dosent limit. There are three note one) this method is acceptable in just in the case $(0,0)$. second) after substitution $x=my^k$ or $y=mx^k$, when you can say It dosent limit that in the limit dosent exist $x$ and $y$ and three) this method is necessary condition not sufficient, I mean with this method just you can show there is no limit not exist of the limit.For your question you should choose $x=my^2$ and substitute it in the limit: $$\frac{my^2y^2}{(my^2)^2+y^4}=\frac{my^4}{m^2y^4+y^4}=\frac{m}{m^2+1}$$ we can see, the limit is dependent to $m$ and so there is no limit in $(0,0)$.
Amin235
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1Hey, sorry I didn't notice this answer until much later due to the lack of refreshing the page. That helps, thanks – Ryan J Mar 19 '16 at 08:37