Show that there exists a d>0, such that f has a limit at (0,0). Find the value of d.

f(x,y) = (x + y)/(x2 + 1) ; e = 0.01

Answer: d = 0.005

(Problem number 63, exercise, 12.2 from Calculus (9th edition by Thomas and Finney.)

- Sep 26th 2008, 07:36 AMtaureau20Using the delta-epsilon definition of the limit -- solve this problem, please!
Show that there exists a d>0, such that f has a limit at (0,0). Find the value of d.

f(x,y) = (x + y)/(x2 + 1) ; e = 0.01

Answer: d = 0.005

(Problem number 63, exercise, 12.2 from Calculus (9th edition by Thomas and Finney.) - Sep 26th 2008, 10:19 AMThePerfectHacker
We will show the limit is zero.

$\displaystyle \left| \frac{x+y}{x^2+1} - 0 \right| = \frac{|x+y|}{x^2+1} \leq |x+y| \leq |x|+|y|$

Now if $\displaystyle 0 < \sqrt{x^2+y^2} < \delta \implies |x|,|y|<\delta$

Therefore $\displaystyle |x|+|y| < 2\delta$.

Thus, choose $\displaystyle \delta = \frac{\epsilon}{2}$.