Prove the Function 

The function considered is given by
$$f(x) = \frac{|x-1| -x+1}{3x-3}$$
which we rewrite as
$$f(x) = \frac{1}3\frac{|x-1|}{x-1}-\frac13\frac{x-1}{x-1}=\frac13\frac{|x-1|}{x-1}-\frac13$$
Now we want to check that this does not admit a limit as $x\to1$. So we look at two sequences, first $x_n=1+\frac1n$ and then $y_n=1-\frac1n$. We will check that $\lim_{n\to\infty} f(x_n)$ and $\lim_{n\to\infty}f(y_n)$ differ, since both $x_n$ and $y_n$ converge to $1$ the conclusion we get is that $f(x)$ does not admit a consistent limit as $x\to1$.

Now:
$$f(x_n)= \frac13\frac{|1+\frac1n-1|}{1+\frac1n-1}-\frac13=\frac13\frac{1/n}{1/n}-\frac13=0$$
and
$$f(y_n) = \frac13\frac{|1-\frac1n-1|}{1-\frac1n-1}-\frac13=\frac13\ \frac{1/n}{-1/n}-\frac13 = -\frac23$$
So we have $\lim_n f(x_n) = \lim_n 0 =0$ and $\lim_n f(y_n) = \lim_n -\frac23 = -\frac23$. These two limits differ, so by the above discussion we find that $f(x)$ does not admit a limit as $x\to1$.