Am I on the right track with this one?
D Djohnst76 New member Joined Feb 27, 2020 Messages 4 Feb 27, 2020 #1 Am I on the right track with this one?
R Romsek Senior Member Joined Nov 16, 2013 Messages 1,362 Feb 28, 2020 #2 not really. What you need to show is [MATH]|ax + b| - |ax-b| = 2ax,~ |x| \leq \dfrac b a[/MATH] and then clearly as \(\displaystyle x \to 0\) eventually \(\displaystyle |x| < \dfrac b a\) and thus [MATH]\lim \limits_{x\to 0} \dfrac{|ax+b|-|ax-b|}{x} = \lim \limits_{x \to 0} \dfrac{2ax}{x} = 2a[/MATH]
not really. What you need to show is [MATH]|ax + b| - |ax-b| = 2ax,~ |x| \leq \dfrac b a[/MATH] and then clearly as \(\displaystyle x \to 0\) eventually \(\displaystyle |x| < \dfrac b a\) and thus [MATH]\lim \limits_{x\to 0} \dfrac{|ax+b|-|ax-b|}{x} = \lim \limits_{x \to 0} \dfrac{2ax}{x} = 2a[/MATH]
D Deleted member 4993 Guest Feb 28, 2020 #3 You can confirm the result above by assuming some nominal values of 'a' and 'b' and observing the behavior of the function around x = 0. The above procedure does not prove that the response #2 is correct - but it shows that it is most probably correct.
You can confirm the result above by assuming some nominal values of 'a' and 'b' and observing the behavior of the function around x = 0. The above procedure does not prove that the response #2 is correct - but it shows that it is most probably correct.
