Answer: D) 5
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Explanation:
If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first
x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator
So, [tex]\frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2[/tex]
Allowing us to say,
[tex]\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\[/tex]
