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\[\large\rm |x-3|=\cases{(x-3),\qquad x\ge3\\ \rm -(x-3),\quad x\lt3}\]
So when we're on the `left side of 3`, Our |x-3| is -(x-3), giving us a function of:\[\large\rm t(x)=2^x-\frac{-(x-3)}{(x-3)}\]Err I guess it would be better to do this with limits. Approach 3 from the left, and seperately approach 3 from the right and see if the two pieces connect. If they don't, we have a jump discontinuity.

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\[\large\rm \lim_{x\to3^-}t(x)=2^{x}-\frac{-(x-3)}{(x-3)}\]Do a cancellation before plugging the 3 in :)
Thanks! @zepdrix
Did you figure out out Princess Jasmine? :o
Yes. I said it was a jump discontinuity. I just needed a starting point.

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