multivarible limits

**lllll** · March 24th 2008, 10:18 PM

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?

**mr fantastic** · March 25th 2008, 03:42 AM

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.

**Moo** · March 25th 2008, 04:24 AM

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.

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