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I have tried to solve $\lim_{x\rightarrow \infty ,y\rightarrow\infty}f(x,y)$, where $f(x,y) = \frac{2x + 3y}{x^2+xy+y^2}$.

Can I define $y = r \sin \theta$ and $x=r\cos \theta$ when $x\rightarrow \infty ,y\rightarrow\infty$ ? Thanks.

Théophile
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Nir Movshovitz
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  • Notice that for $x,y>2$ the denominator is bigger than the numerator. –  May 26 '16 at 15:20
  • There is some mathematical way to prove the limit is 0 ? – Nir Movshovitz May 26 '16 at 15:23
  • You can define x and y that way, but you will need to be careful about the behavior of theta or showing that theta is irrelevant. (I think you know how to go about this, but it is something to be aware of.) – Kitter Catter May 26 '16 at 15:25
  • Im sorry, but i dont understand you. Can you show me what are you talking about ? Thanks. – Nir Movshovitz May 26 '16 at 15:26
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    What does $x\to \infty, y \to \infty$ mean? – zhw. May 26 '16 at 15:27
  • If you do your substitution how does the new limit look? Are you taking the limits as $x\rightarrow\infty$ first or last? Should it matter? Does that affect how your limit on theta shows up? – Kitter Catter May 26 '16 at 15:30
  • It must be zero because of the greater order of the denominator (the denominator grows faster then the numerator). – Ataulfo May 26 '16 at 16:06

2 Answers2

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Write $x=r\cos t, y = r\sin t.$ The expression equals

$$\frac{r(2\cos t+ 3\sin t)}{r^2(\cos^2 t + \cos t\sin t + \sin^2 t)} = \frac{1}{r}\frac{2\cos t+ 3\sin t}{1 + (\sin 2t)/2}.$$

In absolute value the last expression is bounded above by $(1/r)[5/(1/2)] = 10/r.$ This $\to 0$ as $r\to \infty.$

zhw.
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It must be zero because of the greater order of the denominator (the denominator grows faster then the numerator).

Also you can put $$\frac{2x + 3y}{x^2+xy+y^2}=\frac{1}{x}\cdot \frac{2+3\frac yx}{1+\frac yx+(\frac yx)^2}$$ and only if the second factor is of the form $\frac{\infty}{\infty}$ you cannot to be sure that the limit is zero. But in this case $$\lim_{x,y\to \infty} \frac yx=\infty\iff \lim_{x,y\to \infty} \frac xy=0 $$ and you can do the following:

$$\frac{2x + 3y}{x^2+xy+y^2}=\frac{2x + 3y}{(2x+3y)^2-(3x^2+5xy+8y^3)}=\frac {4}{(2x+3y)(4-6x-y)+\frac{29y^2}{2x+3y}}$$ You have $$\frac{29y^2}{2x+3y}=\frac{29}{\frac 1y(2\cdot \frac xy+\frac 3y)}$$ which tends to $\infty$ because $\frac xy$ tend to $0$. It follows that the asked limit is equal to $0$

Ataulfo
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