Need a hint: Multivariable Epsilon-Delta proof

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I'm having troubles prooving this limit

$$\lim_{(x,y)\to (1,2)} {x^2 +1\over y^2 -1} = \frac 23$$

any help would be highly appreciated

Edit1: Thank you, I'll take a look to your comments

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Replace $x$ by $x + 1$, and $y$ by $y + 2$, then we prove:

$$\displaystyle \lim_{(x,y) \to (0,0)} \dfrac{x^2 + 2x + 2}{y^2 + 4y + 3} = \dfrac{2}{3}.$$

First let $\varepsilon > 0$ be given. Choose

$$\delta = \min\left\{\dfrac{1}{4}, \dfrac{31\epsilon}{80}\right\},$$

and set $r = x^2 + y^2$. Then if $\sqrt{r} < \delta$ then $\sqrt{r} < 1$, $r < \sqrt{r}$, and applying the triangle inequality and CS-inequality we have:

$$ \left|\dfrac{x^2 + 2x + 2}{y^2 + 4y + 3} - \dfrac{2}{3}\right| = \left|\dfrac{3x^2 - 2y^2 + 6x - 8y}{3(y^2 + 4y + 3)}\right| < \left|\dfrac{3x^2 + 2y^2 + 10\sqrt{x^2 +y^2}}{3(3 - y^2 - 4|y|)}\right| <\\ <\dfrac{3r + 2r + 10\sqrt{r}}{3(3 - r - 4\sqrt{r})} < \dfrac{15\sqrt{r}}{\dfrac{93}{16}} = \dfrac{80\sqrt{r}}{31} < \dfrac{80\delta}{31} < \varepsilon,$$

and the proof is completed.

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Hint: $$ \frac{x^2+1}{y^2-1} -\frac{2}{3}= \frac{3(x+1)(x-1)+2(2-y)(2+y)}{3(y^2-1)}=\frac{3(x+1)}{3(y^2-1)}(x-1)+\frac{2}{3}\frac{2+y}{y^2-1}(2-y). $$ Now, both $x-1$ and $2-y$ are "as small as we wish" when $(x,y) \to (1,2)$, do you agree? So, can you show that both $$ \frac{3(x+1)}{3(y^2-1)} $$ and $$ \frac{2}{3}\frac{2+y}{y^2-1} $$ remain bounded by some universal constant when $x \approx 1$ and $y \approx 2$? Just use rough estimates of the numerators and of the denominators. Finally, pick $\epsilon>0$ and use these facts to show that $\left| \frac{x^2+1}{y^2-1} -\frac{2}{3} \right|<\epsilon$ when $x\approx 1$ and $y \approx 2$.

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Hint:

To show this using $\epsilon-\delta$ argument, one can consider:

$$\frac{x^2 + 1}{y^2-1} - \frac{2}{3} = \frac{3x^2 +3-2(y^2-1)}{3(y^2-1)} = \frac{3(x^2-1) - 2(y^2 - 4)}{3(y^2-1)}$$

$$\Rightarrow \bigg|\frac{x^2 + 1}{y^2-1} - \frac{2}{3}\bigg| \leq \frac{3|x-1||x+1| + 2|y-2||y+2|}{3|y^2-1|}$$

So you need to show that the right hand side is small, whenever $(x-1)^2 + (y-2)^2$ is small.