2

Consider the function $$ f ( x , y ) = \begin {cases} \frac { 2 x ^ 2 - x y - y ^ 2 } { x ^ 2 - y ^ 2 } & x \ne y \\ \frac 3 2 & x = y \end {cases} $$ on the domain $ D = \left\{ ( x , y ) \in \mathbb R ^ 2 \mid x \ge 1 , y \ge 1 \right\} $.
Determine the region of $ D $ on which $ f $ is continuous.

I calculated that limit does not exist for the individual rational function, does that mean the function is continuous everywhere except when $x=y$ as it must satisfy the definition of $$\lim_{(x,y)\to(a,b)} f(x,y)=(a,b)$$ So when subbing $x=y$ in this case, it gives $\frac{3}{2}$.

Mohsen Shahriari
  • 6,718
yellowpencil
  • 29
  • Welcome to MSE. Please type your questions instead of posting images. Images can't be browsed and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial – saulspatz Aug 22 '20 at 06:06
  • Thanks, I'll make sure to format it the correct way. – yellowpencil Aug 24 '20 at 02:01

1 Answers1

0

$f(x,y)-\frac 3 2 =\frac {(x-y)^{2}} {2(x^{2}-y^{2})}=\frac {x-y} {2(x+y)}$ after some simplification. Hence, for any real number $a \geq 1$ we have $f(x,y)-\frac 3 2 \to 0$ as $(x,y) \to (a,a)$. It follows that $f$ is continuous at $(a,a)$ for any real number $a \geq 1$ . Clearly $f$ is continuous at all other points of its domain. Hence $f$ is continuous everywhere on its domain.

Kavi Rama Murthy
  • 311,013
  • 1
    Hey, thanks for your help. But i dont really understand how did you simplify the equation into that form. In this case, you mean that is continuous everywhere except when x=y ?and shouldn't f(x,y)→3/2 as (x,y)→(a,a) ? Thanks for helping again – yellowpencil Aug 24 '20 at 02:11
  • @yellowpencil Write $2x^2 - xy - y^2$ as $$(x^2 -y^2) + (x^2 -xy) = (x^2 - y^2) + x(x-y).$$ Then for all $(x,y) \in \Bbb R^2$ with $x \neq y$ we have $$f(x,y) = 1 + \frac {x} {x+y}.$$ This implies that for all $(x,y) \in \Bbb R^2$ with $x \neq y$ $$f(x,y) - \frac 3 2 = \frac {x} {x+y} - \frac 1 2 = \frac {x-y} {2(x+y)}.$$ – Anil Bagchi. Aug 24 '20 at 05:18
  • @KaviRamaMurthy sir when $(x,y) \to (a,a)$ then $f(x,y) - \frac 3 2 \to 0 \implies f(x,y) \to \frac 3 2.$ So $f$ is continuous everywhere in $\Bbb R^2.$ – Anil Bagchi. Aug 24 '20 at 05:29
  • @Phibetakappa Thank you. I have corrected my answer. – Kavi Rama Murthy Aug 24 '20 at 05:34