Proving Continuity and Inequalities for a Limit with Two Variables

  • Thread starter Thread starter oahsen
  • Start date Start date
  • Tags Tags
    Limit Variables
oahsen
Messages
58
Reaction score
0

Homework Statement



Let
f(x, y) =((x^2)*y)/(x^4 + y^2) if (x, y) != (0, 0) ,
f(x,y) = 0 if (x, y) = (0, 0) .
a) Is f continuous at (0, 0)? Prove your statement.
b) Show that
-1/2 ≤ f(x, y) ≤1/2
for all (x, y).

I have used the two path test to show that it has not limit at 0,0 hence it is not continuous there. however, ı have no idea what I should do for the b part?Is there a algebratic way to show it or should we take differential etc.?
 
Physics news on Phys.org
Try solving for each inequality separately, and cross multiply to get a perfect square.
 
daveb said:
Try solving for each inequality separately.
Are you referring to the fx and fy (the partial derivatives of f) by saying "for each inequality"?
 
What he means is to find necessary and sufficient conditions on x,y for each of the two inequalities

-1 \leq \frac{2x^2y}{x^4+y^2}

and

1 \geq \frac{2x^2y}{x^4+y^2}

(ie. rearrange them, using "reversible" steps, until you find something that will tell you for which x,y they are satisfied)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top