Find lim x->-3+ x-3/x^2 -9 =?
I'm really confused on this concept.
Note that (x-3)/(x^2-9) = (x-3)/[(x-3)(x+3) = 1/(x+3)
Except, that the quotient is not defined when x = 3 or x = -3, because the denominator is zero.
So, what happens when x gets close to -3? We approach 1/(-3+3) = 1/0, which is not defined. However, we can get as close to -3 as we wish, and still have an answer.
Approaching from the right, as required, consider the values taken on by f = 1/(x+3)
x = -2 f = 1/(-2+3) = 1/1 = 1
x = -2.9 f = 1/(-2.9+3) = 1/.1 = 10
x = -2.99 f = 100
x = -2.999 f = 1000
You can see that the closer x gets to -3, the larger f gets. In fact, f increases without limit, or, as we write:
lim x->-3+ = +oo