At what value of x do you want the limit? Is it at x = 2, where the denominator vanishes?
The numerator is the difference of two squares, andcan be written
(2 + x -4)(2 + x +4)= (x-2)(x+6)
Cancel out the x-2 therms in numerator and denominator anbd you are left with
x + 6. You can easily evaluate that at any value of x, including x=2
evaluate each limit, if it exists
[(2+x)^2 - 16]/ (x-2)
what do I do?
1 answer
1 answer