3. [-/3 Points]
DETAILS
MY NOTES
MARSVECTORCALC6 2.2.006.
You may freely use techniques from one-variable calculus, such as L'Hôpital's rule.
Consider f(x, y).
$$f(x, y) = \begin{cases}
\frac{xy^3}{x^2 + y^6} & \text{if } (x, y) \neq (0, 0) \\
0 & \text{if } (x, y) = (0, 0)
\end{cases}$$
(a) Compute the limit as (x, y) → (0, 0) of f along the path x = 0. (If an answer does not exist, enter DNE.)
(b) Compute the limit as (x, y) → (0, 0) of f along the path x = y³. (If an answer does not exist, enter DNE.)
(c) Show that f is not continuous at (0, 0).
Since the limits as (x, y)→(0, 0) of f along the paths x = 0 and x = y³ are not equal f is not continuous at (0, 0).