# Please explain me how to evaluate the limit of function (x^2-9)/(x+3), without use of derivatives? x goes to -3

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We have to find the value of lim x--> -3 [(x^2-9)/(x+3)]

lim x--> -3 [(x^2-9)/(x+3)]

expand the numerator using a^2 - b^2 = (a - b)(a + b)

=> lim x--> -3 [(x - 3)(x + 3)/(x+3)]

=> lim x--> -3 [(x - 3)]

substitute x = -3

=> -3 - 3

=> -6

**The required value of the limit is -6.**

We notice that if we'll replace x by -3, we'll get an indetermination "0/0" type.

We'll re-write the numerator, that is a difference of 2 squares, as a product.

lim (x^2 - 9)/(x+3) = lim (x - 3)(x + 3)/(x+3)

We'll reduce the fraction by (x+3):

lim (x - 3)(x + 3)/(x+3) = lim (x - 3)

We'll replace x by -3:

lim (x - 3) = -3 - 3 = -6

**The value of the limit is: lim (x^2 - 9)/(x+3) = -6.**