Question - Evaluating Limits of Functions
Solution:
The question displays three functions and asks to evaluate the limits of certain expressions involving these functions.Given:- $$ f: R \rightarrow R $$ with $$ f(x) = x^2 + 5x + 4 $$- $$ g: R \rightarrow R $$ with $$ g(x) = 3x - 3 $$We are asked to find the following limits:a) $$\lim_{x \to -2} [f(x) + g(x)] $$b) $$\lim_{x \to 2} \sqrt{f(x)} - g(x) $$c) $$\lim_{x \to 3} \frac{f(x)}{g(x)} $$Let's solve each one:a) To find the limit of $$ f(x) + g(x) $$ as $$ x $$ approaches $$-2$$, we simply plug $$ -2 $$ into each function and add the results:$$ f(-2) = (-2)^2 + 5(-2) + 4 = 4 - 10 + 4 = -2 $$$$ g(-2) = 3(-2) - 3 = -6 - 3 = -9 $$Now, $$ f(-2) + g(-2) = -2 + (-9) = -11 $$So, $$\lim_{x \to -2} [f(x) + g(x)] = -11$$.b) To find the limit of $$ \sqrt{f(x)} - g(x)$$ as $$x$$ approaches $$2$$, we substitute $$2$$ into each function and then perform the operations, keeping in mind to take the square root of $$f(x)$$ not $$f(x)$$ itself:$$ f(2) = (2)^2 + 5(2) + 4 = 4 + 10 + 4 = 18 $$$$ g(2) = 3(2) - 3 = 6 - 3 = 3 $$Now, the square root of $$ f(2) $$ is $$ \sqrt{18} $$, so:$$ \sqrt{f(2)} - g(2) = \sqrt{18} - 3 $$This result cannot be simplified further without a calculator, so we leave it as $$ \sqrt{18} - 3 $$.Therefore, $$\lim_{x \to 2} \sqrt{f(x)} - g(x) = \sqrt{18} - 3$$.c) To find the limit of $$ \frac{f(x)}{g(x)}$$ as $$x$$ approaches $$3$$, we again substitute $$3$$ into each function and divide the results:$$ f(3) = (3)^2 + 5(3) + 4 = 9 + 15 + 4 = 28 $$$$ g(3) = 3(3) - 3 = 9 - 3 = 6 $$Now, $$ \frac{f(3)}{g(3)} = \frac{28}{6} = \frac{14}{3} $$So, $$\lim_{x \to 3} \frac{f(x)}{g(x)} = \frac{14}{3}$$.To summarize, the limits are as follows:a) $$-11$$b) $$ \sqrt{18} - 3 $$c) $$ \frac{14}{3} $$
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