(a)

If you approach $(0,0)$ along any line $y=mx$, what does $f(x,y)$ approach?

(b)

If you approach $(0,0)$ along the parabola $x=y^2$, what does $f(x,y)$ approach?

(c)

Does $\lim _{(x,y)\rightarrow (0,0)} f(x,y)$ exist?

(a)

What change of coordinates, $u(x,y)$ and $v(x,y)$, can you use to evaluate this limit?

(b)

Evaluate $\lim _{(x,y)\rightarrow (0,0)}f(x,y)$.

(c)

Consider the path $\vec {x}(t) = (t,-t)$. Explain why $\lim _{t\rightarrow 0} f(\vec {x}(t))$ does not exist. Explain why this does not contradict your answer to (b).

Hint: from single variable calculus, what is $\lim _{x\rightarrow 0}\frac {\sin (x)}{x}$?

In this professional problem, you will prove the following statement.

Let $f(x,y) = \displaystyle \dfrac {3x^2 y}{x^2+y^2}$, and let $\varepsilon > 0$ be given. If $\delta = \varepsilon /3$, then $0 < |(x,y) - (0,0)| < \delta$ implies that $|f(x,y)-0|< \varepsilon$, and therefore $\displaystyle \lim _{(x,y) \to (0,0)} f(x,y) = 0$.

Consider the function $$f(x,y) = \dfrac {3x^2 y}{x^2+y^2}.$$ By switching to cylindrical coordinates, you could show the limit of $f$ as $(x,y)\to (0,0)$ is 0. Instead, you will show find limit using the $\varepsilon -\delta$ definition of a limit. Your goal is this: given any $\varepsilon > 0$, you must show there is a $\delta > 0$ so that $$0 < | (x,y) - (0,0) | < \delta \mbox { implies that } |f(x,y)-0| < \varepsilon .$$ For the given point and function, the choice $\delta = \varepsilon /3$ will accomplish this.

Hints: Think back to the limit proofs you did last year. They were separated into Think and Proof sections. The Think portion has already been done for you, inasmuch as the choice of $\delta$ has been made for you. However, you still need to understand why this $\delta$ works in order to provide the proof.

Below is a list of important facts which are necessary to prove this statement. First, convince yourself that they are true on scratch paper. Then, incorporate these facts, with their justifications, into a complete proof.

• $0 < |(x,y)-(0,0)| < \delta \mbox { if and only if } 0 < \sqrt {x^2 + y^2} < \delta$.
• $\displaystyle \left |\dfrac {3x^2 y}{x^2+y^2} - 0 \right | < \varepsilon \mbox { if and only if } \frac {3x^2|y|}{x^2+y^2} < \varepsilon$.
• $\displaystyle \frac {3x^2|y|}{x^2+y^2} \le \frac {3(x^{2}+y^{2})|y|}{x^{2}+y^{2}}$, and therefore $\displaystyle \frac {3x^2|y|}{x^2+y^2} \le 3\sqrt {x^2 + y^2}$.

Focuses: This week, you should focus on these items when writing your solution:

• Organization and Structure: Your proof should be completely self-contained. The facts above are merely suggestions: Their complete and precise statements and justifications should be incorporated into your proof. Write, revise, and rewrite until your entire proof flows together in a logical order. Be concise.
• Explanation: Justify all necessary steps, including those which are not covered by the hints.
• Attention to Mathematical Details: Be very careful with absolute values and strict inequalities. Make sure that your implications follow from each other in the correct order.
• Notation: Use mathematical notation appropriately. Do not use notation in place of english words, or vice versa. Everything, including equations, should be part of a complete (but possibly brief) sentence.