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How do you calculate

$\displaystyle \lim_{(x, y) \to (2, 2)} \dfrac{x^2 + 2y}{x^2 - 2y}$?

My attempt:

I approach the limit by using the line $x = 2$ (because the line passes through $(2, 2)$) Now, I get $\displaystyle \lim_{(x, y) \to (2, 2)} \dfrac{x^2 + 2y}{x^2 - 2y} = \lim_{(x, y) \to (2, 2)} \dfrac{2 + y}{2 - y}$

Now, what is next? If I approach by using the line $y = 2$ or by using the polar coordinates as well, it leads to the same cases. Regards!

Sukardy
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4 Answers4

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I like to have variables go to zero. So, setting $x = a+2, y = b+2$,

$\begin{array}\\ \dfrac{x^2 + 2y}{x^2 - 2y} &=\dfrac{(a+2)^2 + 2(b+2)}{(a+2)^2 - 2(b+2)}\\ &=\dfrac{a^2+4a+4+ 2b+4}{a^2+4a+4 - 2b-4}\\ &=\dfrac{a^2+4a+ 2b+8}{a^2+4a - 2b}\\ \end{array} $

As $a, b \to 0$, the numerator goes to $8$ and the denominator goes to $0$, so the limit does not exist.

marty cohen
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  • Hi Marty. Happy New Year. The limit fails to exist, but not simply because the denominator approaches $0$. Note that it can approach $0^+$ or $0^-$ – Mark Viola Jan 07 '18 at 04:43
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    Yes, it is also possible to go directly for $f(2,2+b)=-1-\frac 8{2b}$ which has a simpler form and shows the $\pm\infty$ divergence. (i.e. $a=0$). But anyway, the method is fine, so (+1). – zwim Jan 07 '18 at 04:45
1

$f(x,y) = \dfrac{x^2+2y}{x^2-2y}$.

Consider 2 sequences z_n :=(x_n,y_n) converging to (2,2).

1) Let $x_n= 2 +1/n$, $y_n= 2+2/n$.

$f(x_n,y_n) = $

$\dfrac{ 4+4/n +1/n^2 +4 +4/n}{4 +4/n +1/n^2 - 4 - 4/n}=$

$\dfrac{8 +8/n +1/n^2}{1/n^2} =$

$n^2(8+8/n +1/n^2) = $

$8n^2 +8n +1.$

$\lim_{n.\rightarrow \infty}f(x_n,y_n) = +\infty.$

2) Now let

$x_n = 2 - 1/n$; $y_n = 2 +2/n$ ;

$f(x_n,y_n) = $

$\dfrac{4 -4/n +1/n^2 + 4 +4/n}{4 -4/n +1/n^2 -4 - 4/n}=$

$\dfrac{8 +1/n^2}{ -8/n +1/n^2}=$

$-\dfrac{8n +1/n}{ 8 - 1/n}.$

$\lim_{n \rightarrow \infty} f(x_n,y_n)= -\infty.$

Hence the limit does not exist.

Peter Szilas
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0

It is obvious that the limit doesn't exist because when you plug $(2,2)$ the numerator remains positive while the denominator tends to $0$.

0

The limit does not exists, indeed for $x=2+t$ and $y=2-t$ for $t\to0$

$$\dfrac{x^2 + 2y}{x^2 - 2y}=\dfrac{(2+t)^2 + 2(2-t)}{(2+t)^2 - 2(2-t)}=\dfrac{t^2+2t+8}{t^2+6t}=\dfrac{t+2+\frac8t}{t+6}\to\begin{cases}+\infty \quad t\to0^+\\-\infty \quad t\to0^- \end{cases}$$

user
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