How do you find the limit of #((t^2) + 2)/(t-4)# as t approaches 4?

1 Answer
Aug 25, 2016

Limit does not exist

Explanation:

start by plugging in 4

#(4^2 + 2)/(4-4)# so we see that this goes to #oo# at #t = 4#

the numerator is always positive

....whereas the denominator is negative for #t = 4 - delta# and positive for #t = 4 + delta#, where #0 < delta " << " 1# so we have a two-sided limit

#lim_(t to 4^-) ((t^2) + 2)/(t-4) = -oo#

#lim_(t to 4^+) ((t^2) + 2)/(t-4) = oo#

Therefore limit does not exist