Continuity of a bivariate function

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $$f(x,y) = \begin{cases} \frac{x^2 y^2}{x^4 + y^4}, & \text{ }\text{(x,y)} \neq (0,0) \\ 0, & \text{ }\text{(x,y)} = (0,0) \end{cases} .$$

Show that $\frac{df}{dx} (0,0)$ exists, and $f$ is not continuous at $(0,0)$.

-

1
Write down the limit defining the partial derivative. What do you get? Compute the limit of $f$ along the path $x=y$ and compare with that value of $f(0,0)$. What do you get? – David Mitra Apr 2 '12 at 23:27

1 Answer

For existence of the partial derivative:

By definition, $f_x(0,0)$ is the limit $$\lim_{h\rightarrow 0}{f(h,0)-f(0,0)\over h }.$$ Show that this limit exists for your particular function.

For continuity:

Note that,
$$\lim_{x\rightarrow 0} f(x,x) =\lim_{x\rightarrow 0}{x^4\over2x^4}={1\over2}.$$ Keeping this in mind and the fact that $f(0,0)=0$, what can you say about the continuity of $f$ at $(0,0)$?

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9151668548583984, "perplexity_flag": "head"}