Use the $\varepsilon - \delta$ definition of the limit to verify that $\lim_{(x,y)\to(2,5)} xy = 10$.

Question

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Question:

Use the $\varepsilon - \delta$ definition of the limit to verify that $\lim\limits_{(x,y)\to(2,5)}xy = 10$. Hint given: $xy − 10 = (x − 2)(y − 5) + 5(x − 2) + 2(y − 5).$

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My solution

$\forall \space\varepsilon\gt0\space\space\exists\space\delta(\varepsilon)\gt0:|xy-10|\lt\varepsilon$ where $(x,y)\in D$, whenever $0\lt\sqrt{(x-2)^2+(y-5)^2}\lt\delta$

Note:

$(x-2)^2 \le (x-2)^2+(y-5)^2$ $\implies |x-2|\le\sqrt{(x-2)^2+(y-5)^2}$, similarly, $|y-5|\le\sqrt{(x-2)^2+(y-5)^2}$

So if $(x,y)\in D\space$ and $\space0\lt\sqrt{(x-2)^2+(y-5)^2}\lt\delta$. We choose $ \delta :=\frac{-7+\sqrt{49+4\varepsilon}}2$.

We obtain...

$\begin{aligned}|xy-10|&=|(x − 2)(y − 5) + 5(x − 2) + 2(y − 5)|\\&\le|x − 2||y − 5|+ 5|x − 2| + 2|y − 5|\\&\le \left((x-2)^2 +(y-5)^2\right) +5\sqrt{(x-2)^2+(y-5)^2}+2\sqrt{(x-2)^2+(y-5)^2}\\&=\left(\sqrt{(x-2)^2+(y-5)^2}\right)^2+7\sqrt{(x-2)^2+(y-5)^2}\\&\lt \delta^2+7\delta\\&=\frac{49\pm\sqrt{49-4\varepsilon}+49+4\varepsilon}4+\frac{-98\pm14\sqrt{49+4\varepsilon}}4\\&=\frac{4\varepsilon}4\\&=\varepsilon\end{aligned}$

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Comments

This is a question on my Analysis 2 module. Would be great if someone could check my solution. Also, I realise you have to use a minimum for $\delta$, but unsure on how to unpack it when writing the solution, would be great if anyone can make my understanding of this clearer.

Answer 1

Note that since $\epsilon >0$, $\sqrt{ 49 + 4\epsilon } > \sqrt{49} = 7$. So your $\delta$ is negative.

To remedy this, you can choose $$ \delta = \frac{-7+\sqrt {49+4\epsilon}}{4},$$ which is positive and $\delta^2 + 7\delta = \epsilon$.

Another method: note that you obtained

$$ |xy-10| \le \delta^2 + 7\delta$$

and you need that $\delta ^2 + 7\delta <\epsilon$. For example, if you choose $\delta <1$, then

$$ \delta ^2 + 7\delta < \delta + 7\delta.$$

Then if you choose $ \epsilon = \min\{ 1, \delta/8\}$, you have $$|xy-10| \le \delta^2 + 7\delta<\delta + 7\delta < \epsilon. $$

Answer 2

First observe that: $$ |xy-10|=|xy-2y+2y+10|\le |y||x-2|+2|y-5| $$ Then set the first restriction: $\delta_1=1$ $$ |y-5|\le \big((x-2)^2+(y-5)^2)\big)^{½}<1=\delta_1, $$ which implies, that $$ |y|\le |y-5|+5<6, $$ and hence, if $\big((x-2)^2+(y-5)^2)\big)^{½}<1$, then $$ |xy-10|=|xy-2y+2y+10|\le |y||x-2|+2|y-5|\le 6|x-2|+2|y-5|\\ \le (6^2+2^2)^{1/2}\big((x-2)^2+(y-5)^2)\big)^{½} \le 7\big((x-2)^2+(y-5)^2)\big)^{½} $$ Hence, if $$ \big((x-2)^2+(y-5)^2)\big)^{½}<\delta=\min\{\delta_1,\varepsilon/7\}=\min\{1,\varepsilon/7\}, $$ then $$ |xy-25|\le 8\big((x-2)^2+(y-5)^2)\big)^{½}<\varepsilon. $$