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Evaluate a 0/0 limit using L'Hopital's Rule

Consider the function f(x)=x2−9x−3.f(x) = \dfrac{x^2 - 9}{x - 3}. Evaluate lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3} using L'Hopital's Rule. First confirm that direct substitution produces an indeterminate form of type 0/00/0, then apply the rule to find the limit.

Hint

Substitute x=3x = 3 into the numerator and denominator separately to confirm both approach 0. Then differentiate the numerator and denominator separately (do not use the quotient rule), and take the limit of the resulting ratio.

Step-by-step solution

Substituting x=3x = 3 gives numerator 32−9=03^2 - 9 = 0 and denominator 3−3=03 - 3 = 0, so the limit has the indeterminate form 0/00/0. This confirms L'Hopital's Rule applies here.

Differentiate the numerator and denominator separately: the derivative of x2−9x^2 - 9 is 2x2x, and the derivative of x−3x - 3 is 11. So the limit becomes lim⁡x→32x1=2(3)=6.\lim_{x \to 3} \frac{2x}{1} = 2(3) = 6.

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