Answer: B) 4/3
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Work Shown:
[tex]\displaystyle \lim_{x \to 3} \frac{f^2(x)-4}{f^2(x)-f(x)-2}\\\\\\\displaystyle \lim_{x \to 3} \frac{(f(x)-2)(f(x)+2)}{(f(x)-2)(f(x)+1)}\\\\\\\displaystyle \lim_{x \to 3} \frac{f(x)+2}{f(x)+1}\\\\\\\displaystyle \frac{f(3)+2}{f(3)+1}\\\\\\\displaystyle \frac{2+2}{2+1}\\\\\\\displaystyle \frac{4}{3}\\\\\\[/tex]
Notes:
In step 2, I used the difference of squares rule to factor the numerator. The denominator can be factored through trial and error. The key here is the f(x)-2 terms that show up in each. Those factors cancel in step 3.
Afterward, we apply the substitution rule, [tex]\displaystyle \lim_{x \to a} f(x) = f(a)[/tex], so basically I plugged in x = 3. Then evaluated f(3) = 2 due to the point (3,2).
