Evaluating Limits of Functions
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) \( \frac{14}{3} \)
