Limit of an absolute function

the question is: Limit as X aproaches 4, absolute value (16-t^2)/(4-t)
I'm not entirely sure how to go about starting this, thanks

and IF you meant | 16 - t² | / [ 4 - t ] , then the limit does not exist..

8 from one side while -8 from the other...clarity is all important in mathematics

16 - t² = 4² - t² = (4 + t)(4 - t)
-->

So your expression can be simplified because the (4 - t)'s cancel, leaving

|4 + t|

This is what the graph will look like everywhere except @ t = 4 (when 4 - t = 0), because at that point this expression is UNDEFINED (because you cannot divide by 0)...however, it will approach this point because everywhere else (i.e. infinitely close to t = 4) it will be |4 + t|

So now it's just a matter of plugging in t = 4
-->

|4 + 4| = 8

(16-t^2)/(4-t) can be written as:

(4-t) * (4+t) / (4-t)

Since t is approaching 4 (not equal to 4) we can cancel 4-t from numerator and denominator. which give

(16-t^2)/(4-t) = 4+t

When t approach 4, (16-t^2)/(4-t) will approach 4+t = 8.

(16-t^2)/(4-t) = (4+t)(4-t)/(4-t) = 4+t
So, the limit = 4+4 = 8

lim(t->4) |(16-t^2)/(4-t)|
= lim(x->4) |(4 - t)(4 + t) / (4-t)|
= |4 + t| = 8