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Find the limit of $\frac{4x^4 + 5y^4}{x^2 + y^2}$ as $(x,y)\to (0,0)$.

Which method do I use to find the limit of that? I tried paths but the limits all came out to be $0$... (as a side question, when do you stop trying paths? I mean there are so many ways to try out when $x$ approaches $0$. You can try $y=0$, $y=x$, $y=x^2$, $y=mx$, and so many more ways. After you get like $0$ for 4 limits, do you just stop there and assume to try another method?) (Also, when I try different ways for paths, will the limits always be either $0$ or a finite number and never DNE?)

Thank you

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And what is wrong with 0? It is a respectable member of $\mathbb{R}$ too, you know... –  vonbrand Apr 13 '13 at 20:23
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6 Answers

The euclidean norm is $$||(x,y)||=\sqrt{x^2+y^2}$$ and we have $$x^4\leq||(x,y)||^4\quad;\quad y^4\leq||(x,y)||^4$$ so we have $$0\leq\frac{4x^4 + 5y^4}{x^2 + y^2}\leq9||(x,y)||^2\to_{(x,y)\to(0,0)}0$$

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@PeterTamaroff: No, there's a norm squared in the denominator. –  Javier Badia Apr 14 '13 at 2:25
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@JavierBadia Ooops. –  Pedro Tamaroff Apr 14 '13 at 2:25
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There is a nearly universal strategy when the denominator is $x^2+y^2$ or a close relative. Let $x=r\cos \theta$, $y=r\sin\theta$.

Here the bottom becomes $r^2$, and the top is $4r^4\cos^4\theta+5r^5\sin^5\theta$. Divide. We get $4r^2\cos^4\theta+5r^3\sin^5\theta$. The trigonometric functions are bounded, so the limit is $0$.

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In principle, you can never stop looking for new paths and may have to be creative. However, you are right to suspect that the limit is indeed $0$. To show this, you better prove that for each $\epsilon>0$ there exists $\delta>0$ such that $\sqrt{x^2+y^2}<\delta$ implies $|f(x,y)|<\epsilon$.

For your second question: Virtually anything can happen along different paths - different limits, divergence to infinity, and and of course proper divergence (think of zigzagging between two paths with different limits - the zigzag path will not have a limit).

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good advice ..+1 –  Halil Duru Apr 13 '13 at 20:20
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Is this what you mean?

$$\underset{x,y\rightarrow0}{\lim}\frac{4x^4+5y^4}{x^2+y^2}=$$

$$=\underset{x,y\rightarrow0}{\lim}\frac{4x^4}{x^2+y^2}+\underset{x,y\rightarrow0}{\lim}\frac{5y^4}{x^2+y^2}=$$ $$=\underset{x\rightarrow0}{\lim}\frac{4x^4}{2x^2}+\underset{y\rightarrow0}{\lim}\frac{5y^4}{2y^2}=$$ $$=\underset{x\rightarrow0}{\lim}2x^2+\underset{y\rightarrow0}{\lim}\frac{5}{2}y^2=0+0=0$$

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You need to justify some passages to limit their operations here. –  Elias Apr 13 '13 at 19:24
    
Well this is the intuitive approach. –  Ambesh Apr 13 '13 at 19:27
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But its construction is very good. The justifications just saw that can be fitted perfectly in the middle of his solution. –  Elias Apr 13 '13 at 19:33
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The standard procedure is indeed changing the coordinates from Cartesian to Polar. Your limit:

$$ \lim_{(x, y)\to(0, 0)} \frac{4x^4 + 5y^4}{x^2 + y^2} $$

Will become (with $x=r\cos(\theta)$ and $y=r\sin(\theta)$ and using the fact that $r\to0$ when $(x, y)\to(0, 0)$): $$ \lim_{r\to0} \frac{4r^4\cos(\theta)^4 + 5r^4\sin(0)^4}{r^2\cos(\theta)^2 + r^2\sin(\theta)^2} $$

which ONLY exists if it is the same for all values of $\theta$ (the limit does not vary with $\theta$).

Using: $\sin^2(\theta)+\cos^2(\theta)=1$ your limit will further reduce to:

$$\lim_{r\to0} 4r^2 \to 0 \text{ for all } \theta \text{; therefore the limit exists.}$$

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Well , let $x^2+y^2=r^2$ then $$\frac{4x^4 + 5y^4}{x^2 + y^2}\leq\frac {5 r^4}{r^2} =5r^2 \to 0$$ as $r \to 0$ $\hspace{99mm} \blacksquare$

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$\frac{5r^4}{r^2}$. –  Eric Jablow Apr 14 '13 at 1:06
    
this version better i guess --thank you... –  Halil Duru Apr 14 '13 at 5:15
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