Question

examine the behavior of (f(x,y)=\frac{3x^{2}y}{7x^{4}+y^{2}}) as ((x,y)) approaches ((0,0)). (a) taking a linear approach to the origin along the (y -)axis, we find (lim_{(x,y)\to(0,0),x = 0}left(\frac{3x^{2}y}{7x^{4}+y^{2}}
ight)=). (b) taking a linear approach to the origin along the (x -)axis, we find (lim_{(x,y)\to(0,0),y = 0}left(\frac{3x^{2}y}{7x^{4}+y^{2}}
ight)=). (c) taking a linear approach to the origin along a line (y=mx) with (m
eq0) we find (lim_{(x,y)\to(0,0),y = mx}left(\frac{3x^{2}y}{7x^{4}+y^{2}}
ight)=lim_{x\to0}left(
ight)=). thought question: do your answers to parts (a)-(c) allow you to conclude that the limit exists?

Explanation:

Step1: Find limit along y - axis (a)

Set \(x = 0\). Then \(f(x,y)=\frac{3x^{2}y}{7x^{4}+y^{2}}\) becomes \(\lim_{(x,y)\to(0,0),x = 0}\frac{3x^{2}y}{7x^{4}+y^{2}}=\lim_{y\to0}\frac{0}{y^{2}} = 0\).

Step2: Find limit along x - axis (b)

Set \(y = 0\). Then \(f(x,y)=\frac{3x^{2}y}{7x^{4}+y^{2}}\) becomes \(\lim_{(x,y)\to(0,0),y = 0}\frac{3x^{2}y}{7x^{4}+y^{2}}=\lim_{x\to0}\frac{0}{7x^{4}} = 0\).

Step3: Find limit along \(y=mx\) (c)

Substitute \(y = mx\) into \(f(x,y)\):
\[

$$\begin{align*} \lim_{(x,y)\to(0,0),y = mx}\frac{3x^{2}y}{7x^{4}+y^{2}}&=\lim_{x\to0}\frac{3x^{2}(mx)}{7x^{4}+(mx)^{2}}\\ &=\lim_{x\to0}\frac{3mx^{3}}{7x^{4}+m^{2}x^{2}}\\ &=\lim_{x\to0}\frac{3mx^{3}}{x^{2}(7x^{2}+m^{2})}\\ &=\lim_{x\to0}\frac{3mx}{7x^{2}+m^{2}}\\ &= 0 \end{align*}$$

\]

Since the limit of \(f(x,y)=\frac{3x^{2}y}{7x^{4}+y^{2}}\) as \((x,y)\to(0,0)\) is \(0\) along the \(y - axis\), \(x - axis\) and the line \(y = mx\), we can conclude that \(\lim_{(x,y)\to(0,0)}\frac{3x^{2}y}{7x^{4}+y^{2}}=0\).

Answer:

The limit exists and is equal to \(0\).