The discussion revolves around the concept of differentiability in vector calculus, specifically focusing on a problem involving the limit definition of differentiability and vector functions. The original poster presents a mathematical expression and seeks guidance on how to manipulate it for analysis.
Some participants have provided guidance on how to approach the problem by suggesting specific paths for \(\vec{h}\) and how to categorize the terms in the expression. There is an ongoing clarification about the notation used, indicating that participants are actively engaging with the material.
There is a mention of the need to consider limits from multiple directions, which may imply constraints on the approach to the problem. The original poster's inquiry about notation suggests a potential gap in understanding that could affect their analysis.
HallsofIvy said:I recommend looking at [itex]\vec{h}[/itex] going to 0 along the coordinate axes separately.
If [itex]\vec{h}= <h, 0, 0>[/itex] then [itex]\vec{c}\times\vec{h}[/itex]= <0, -c_yh, c_zh>[/itex], [itex]||\vec{a}+ \vec{h}||^2\vec{c}= (a_x^2h^2+ 2a_xh+ h^2+ a_y^2+ a_z^2)\vec{c}[/itex], and [itex]||\vec{a}||^2\vec{c}= (a_x^2+ a_y^2+ a_z^2)\vec{c}[/itex], so that [itex]||\vec{a}+ \vec{h}||^2\vec{c}- ||\vec{h}||^2\vec{c}= (2a_xh+ h^2}\vec{c}[/itex]. Separate that into parts that are linear in h (D) and non-linear in h (E).
Do the same for [itex]\vec{h}= < 0, h, 0>[/itex] and [itex]\vec{h}= < 0, 0, h>[/itex]. Show that [itex]\displaytype\lim_{h\to 0}E/h= 0[/itex] from all three directions.
Ted123 said:What do you mean by the notation < ... > ?