To The Absolute Limit

To solve this problem, we consider the two cases for absolute value. If $x - 1 < 0$ , we can replace $|x - 1|$ with $-(x-1)$ . We can see that as x approaches 1 from the left, x - 1 approaches 0 from the left. Therefore,

$\lim_{x \to 1^-} \frac{|x-1|}{x-1} = \lim_{x \to 1^-} \frac{-(x-1)}{x-1} = -1$

Similarly, for the case where $x - 1 > 0$ , we have $|x - 1| = x - 1$ .

$\lim_{x \to 1^+} \frac{|x-1|}{x-1} = \lim_{x \to 1^+} \frac{x-1}{x-1} = 1$

Since the left hand limit does not equal the right hand limit, the limit does not exist.

L'Hospital's principle can be used, I think?

$\frac {dy}{dx} |x|$ does not exist for $x=0$

Let $u = x-1$

Finding the new limit:

$x \rightarrow 1 \Rightarrow u \rightarrow 0$

From knowing that the derivative of the modulus of some variable doesn't exist when the variable is equal to zero; the limit equally doesn't exist. L'Hospital's principle would suggest that the limit also doesn't exist?