of What?Show that the limit of as x approaches 1 does not exist. Show the complete notation in this problem before solving and proving.
Are you registered for the mind-reading class in this semester?the OP means the function f(x) = |x-1|.
Of course the OP means the function f(x) = |x-1|.
I think that professor Steven mixed between the continuity and differentiability of [imath]|x - 1|[/imath] at [imath]x = 1[/imath]. Yeah, this function is continuous at [imath]x = 1[/imath] because its limit exists as [imath]x \rightarrow 1[/imath], but it isn't differentiable at [imath]x = 1[/imath].The limit of |x - 1| as x approaches 1 exists, and it equals 0.
Yes, you are correct! BTW, I failed the mind reading class.I think that professor Steven mixed between the continuity and differentiability of [imath]|x - 1|[/imath] at [imath]x = 1[/imath]. Yeah, this function is continuous at [imath]x = 1[/imath] because its limit exists as [imath]x \rightarrow 1[/imath], but it isn't differentiable at [imath]x = 1[/imath].
You didn't fail completely. The probability the OP meant [imath]\ \displaystyle \lim_{x\rightarrow 1}\frac{1}{x - 1} \ [/imath] is very high.Yes, you are correct! BTW, I failed the mind reading class.
OP meant limx→11x−1 \ \displaystyle \lim_{x\rightarrow 1}\frac{1}{x - 1} \ x→1limx−11 is very high. (response #8 above}
I have just guessed this conclusion by comparing post #1 with post #3 written by Pro. Steven.How did you come you this conclusion from the Original Post?
Of course, guessing what the actual problem is, is different from guessing a possible solution to a problem and checking whether it works.I have just guessed this conclusion by comparing post #1 with post #3 written by Pro. Steven.
Note: If we don't make guesses, non of the differential equations in the world would have been solved!