1. ## multivarible limits

Problems:
1) $\lim_{(x,y) \rightarrow (0,0)} \frac{x^2 +sin^2(y)}{2x^2 +y^2}$

2) $\lim_{(x,y) \rightarrow (0,0)} \frac{x^4-y^4}{x^2+y^2}$

3) $\lim_{(x,y) \rightarrow (0,0)} \frac{xy^4}{x^2+2y^2}$

Solutions:
1) $x=0 \ \ \mbox{then} \ \ f(0,y) = \frac{0+sin^2(y)}{0+y^2} = \frac{sin^2(y)}{y^2}$
$y=0 \ \ \mbox{then} \ \ f(x,0) = \frac{x^2+0}{2x^2+0} = \frac{x^2}{2x^2} = 1$

Since the limits aren't the same the limit D.N.E.

2) $x=0 \ \ \mbox{then} \ \ f(0,y) = \frac{0-y^4}{0+y^2} = -y^2$
$y=0 \ \ \mbox{then} \ \ f(x,0) = \frac{x^4-0}{x^2+0} = x^2$

Since the limits aren't the same the limit D.N.E.

3) $y=mx, \ \ x=y^4$
$f(x,y) = f(x,mx) = \frac{x(mx)^4}{x^2+(mx)^8} = \frac{xm^4x^4}{x^2+m^8x^8} = \frac{m^4x^3}{1+m^8x^6}$

$f(x,y) = f(y^4,y) = \frac{y^4 \times y^4}{(y^4)^2+y^8} = \frac{y^8}{y^8+y^8} = \frac{1}{2}$

Since the limits aren't the same the limit D.N.E.

are these correct?

2. Originally Posted by lllll
Problems:
1) $\lim_{(x,y) \rightarrow (0,0)} \frac{x^2 +sin^2(y)}{2x^2 +y^2}$

2) $\lim_{(x,y) \rightarrow (0,0)} \frac{x^4-y^4}{x^2+y^2}$

3) $\lim_{(x,y) \rightarrow (0,0)} \frac{xy^4}{x^2+2y^2}$

Solutions:
1) $x=0 \ \ \mbox{then} \ \ f(0,y) = \frac{0+sin^2(y)}{0+y^2} = \frac{sin^2(y)}{y^2}$ Mr F says: And ${\color{red}\lim_{y \rightarrow 0} \frac{sin^2(y)}{y^2} = 1}$.
$y=0 \ \ \mbox{then} \ \ f(x,0) = \frac{x^2+0}{2x^2+0} = \frac{x^2}{2x^2} = 1$

Since the limits aren't the same the limit D.N.E. Mr F says: Unfortunately they are the same.

2) $x=0 \ \ \mbox{then} \ \ f(0,y) = \frac{0-y^4}{0+y^2} = -y^2$ Mr F says: And ${\color{red}\lim_{y \rightarrow 0} (-y^2) = 0}$.
$y=0 \ \ \mbox{then} \ \ f(x,0) = \frac{x^4-0}{x^2+0} = x^2$ Mr F says: And ${\color{red}\lim_{x \rightarrow 0} x^2 = 0}$.

Since the limits aren't the same the limit D.N.E. Mr F says: Unfortunately they are the same.

3) $y=mx, \ \ x=y^4$
$f(x,y) = f(x,mx) = \frac{x(mx)^4}{x^2+(mx)^8} = \frac{xm^4x^4}{x^2+m^8x^8} = \frac{m^4x^3}{1+m^8x^6}$

$f(x,y) = f(y^4,y) = \frac{y^4 \times y^4}{(y^4)^2+y^8} = \frac{y^8}{y^8+y^8} = \frac{1}{2}$

Since the limits aren't the same the limit D.N.E.

are these correct?
3) is good. You need to have another think about 1) and 2). Please post if after that think you're still stuck.

3. Hello,

For the second one, you can simplify :

$x^4-y^4=(x^2+y^2)(x^2-y^2)$

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

Which tends to 0, for all directions of x and y.