Question

in this problem we show that the function (f(x,y)=\frac{3x^{2}-y}{x^{2}+y}) does not have a limit as ((x,y)\to(0,0)). (a) suppose that we consider ((x,y)\to(0,0)) along the curve (y = 2x^{2}). find the limit in this case: (lim_{(x,2x^{2})\to(0,0)}\frac{3x^{2}-y}{x^{2}+y}=) (b) now consider ((x,y)\to(0,0)) along the curve (y = 3x^{2}). find the limit in this case: (lim_{(x,3x^{2})\to(0,0)}\frac{3x^{2}-y}{x^{2}+y}=) (c) note that the results from (a) and (b) indicate that (f) has no limit as ((x,y)\to(0,0)) (be sure you can explain why!). to show this more generally, consider ((x,y)\to(0,0)) along the curve (y=mx^{2}) for arbitrary (m). find the limit in this case: (lim_{(x,mx^{2})\to(0,0)}\frac{3x^{2}-y}{x^{2}+y}=) be sure that you can explain how this result also indicates that (f) has no limit as ((x,y)\to(0,0)).

Explanation:

Step1: Substitute $y = 2x^{2}$ into the function

Substitute $y = 2x^{2}$ into $f(x,y)=\frac{3x^{2}-y}{x^{2}+y}$.
\[

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

\]

Step2: Simplify the expression

Simplify the fraction $\frac{3x^{2}-2x^{2}}{x^{2}+2x^{2}}$.
\[

$$\begin{align*} \lim_{x\to0}\frac{3x^{2}-2x^{2}}{x^{2}+2x^{2}}&=\lim_{x\to0}\frac{x^{2}}{3x^{2}}\\ &=\lim_{x\to0}\frac{1}{3}\\ &=\frac{1}{3} \end{align*}$$

\]

Step3: Substitute $y = 3x^{2}$ into the function

Substitute $y = 3x^{2}$ into $f(x,y)=\frac{3x^{2}-y}{x^{2}+y}$.
\[

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

\]

Step4: Simplify the new - expression

Simplify the fraction $\frac{3x^{2}-3x^{2}}{x^{2}+3x^{2}}$.
\[

$$\begin{align*} \lim_{x\to0}\frac{3x^{2}-3x^{2}}{x^{2}+3x^{2}}&=\lim_{x\to0}\frac{0}{4x^{2}}\\ &= 0 \end{align*}$$

\]

Step5: Generalize for $y=mx^{2}$

Substitute $y = mx^{2}$ into $f(x,y)=\frac{3x^{2}-y}{x^{2}+y}$.
\[

$$\begin{align*} \lim_{(x,mx^{2})\to(0,0)}\frac{3x^{2}-y}{x^{2}+y}&=\lim_{x\to0}\frac{3x^{2}-mx^{2}}{x^{2}+mx^{2}}\\ &=\lim_{x\to0}\frac{(3 - m)x^{2}}{(1 + m)x^{2}}\\ &=\frac{3 - m}{1 + m},m eq - 1 \end{align*}$$

\]
Since the limits are different for different paths ($y = 2x^{2}$ gives $\frac{1}{3}$ and $y = 3x^{2}$ gives $0$), the limit $\lim_{(x,y)\to(0,0)}\frac{3x^{2}-y}{x^{2}+y}$ does not exist.

Answer:

The limit $\lim_{(x,y)\to(0,0)}\frac{3x^{2}-y}{x^{2}+y}$ does not exist because the limits along different paths $y = 2x^{2}$ and $y = 3x^{2}$ are different ($\frac{1}{3}$ and $0$ respectively).