Question 9 Let limx→3f(x)=5\displaystyle \lim_{x \to 3} f(x) = 5 and limx→3g(x)=−2\displaystyle \lim_{x \to 3} g(x) = -2. Evaluate the following limit using limit laws: limx→3(f(x)2+3g(x)g(x)−f(x))\lim_{x \to 3} \left( \frac{f(x)^2 + 3g(x)}{g(x) - f(x)} \right) Justify each step using properties of limits. Show solutionHide solution+Question 9 - Solution We are given: limx→3f(x)=5,limx→3g(x)=−2\lim_{x \to 3} f(x) = 5, \qquad \lim_{x \to 3} g(x) = -2 We need to evaluate: limx→3(f(x)2+3g(x)g(x)−f(x))\lim_{x \to 3} \left( \frac{f(x)^2 + 3g(x)}{g(x) - f(x)} \right) Step 1: Use limit laws in numerator and denominator. Numerator: limx→3f(x)2=(limx→3f(x))2=52=25\lim_{x \to 3} f(x)^2 = \left( \lim_{x \to 3} f(x) \right)^2 = 5^2 = 25 limx→33g(x)=3⋅limx→3g(x)=3⋅(−2)=−6\lim_{x \to 3} 3g(x) = 3 \cdot \lim_{x \to 3} g(x) = 3 \cdot (-2) = -6 So, limx→3(f(x)2+3g(x))=25−6=19\text{So, } \lim_{x \to 3} (f(x)^2 + 3g(x)) = 25 - 6 = 19 Denominator: limx→3(g(x)−f(x))=limx→3g(x)−limx→3f(x)=−2−5=−7\lim_{x \to 3} (g(x) - f(x)) = \lim_{x \to 3} g(x) - \lim_{x \to 3} f(x) = -2 - 5 = -7 Step 2: Final Answer limx→3(f(x)2+3g(x)g(x)−f(x))=19−7=−197\lim_{x \to 3} \left( \frac{f(x)^2 + 3g(x)}{g(x) - f(x)} \right) = \frac{19}{-7} = \boxed{-\frac{19}{7}}
