It depends on how t is approaching 2.
In fact, if you consider the right limit, you have
[tex] \displaystyle \lim_{t \to 2^+} \dfrac{2-t}{|t-2|} [/tex]
But since t is greater than 2, t-2 is positive, and thus |t-2| = t-2.
So, the limit becomes
[tex] \displaystyle \lim_{t \to 2^+} \dfrac{2-t}{t-2} = \lim_{t \to 2^+} -1 = -1 [/tex]
On the other hand, if you consider the left limit, you have
[tex] \displaystyle \lim_{t \to 2^-} \dfrac{2-t}{|t-2|} [/tex]
But since t is less than 2, t-2 is negative, and thus |t-2| = -t+2.
So, the limit becomes
[tex] \displaystyle \lim_{t \to 2^+} \dfrac{2-t}{-t+2} = \lim_{t \to 2^+} 1 = 1 [/tex]
So, this limit does not exist, because the left and right limits exist but are not the same.
