1. **State the problem:** We are given the function $f(x) = \sqrt{3x - 2}$ and asked to find: - The value of the function at $x = 1$. - The limit of the function as $x \to 1$. 2. **Formula and rules:** The function is defined as $f(x) = \sqrt{3x - 2}$. To find the value at $x=1$, substitute directly. To find the limit as $x \to 1$, we can use three methods: direct substitution, graphing, and tabulation. 3. **Evaluate the function at $x=1$:** $$f(1) = \sqrt{3(1) - 2} = \sqrt{3 - 2} = \sqrt{1} = 1$$ 4. **Find the limit as $x \to 1$ by direct substitution:** Since the function is continuous at $x=1$ (the expression under the square root is positive), the limit is the same as the function value: $$\lim_{x \to 1} \sqrt{3x - 2} = \sqrt{3(1) - 2} = 1$$ 5. **Find the limit by graphing:** The graph of $f(x) = \sqrt{3x - 2}$ is the upper half of a parabola shifted and scaled. Near $x=1$, the graph approaches the point $(1,1)$ smoothly, confirming the limit is 1. 6. **Find the limit by tabulation:** Create a table of values approaching $x=1$ from left and right: | $x$ | $f(x) = \sqrt{3x - 2}$ | |-----|-----------------------| | 0.9 | $\sqrt{3(0.9)-2} = \sqrt{2.7-2} = \sqrt{0.7} \approx 0.837$ | | 0.99| $\sqrt{3(0.99)-2} = \sqrt{2.97-2} = \sqrt{0.97} \approx 0.985$ | | 1.0 | 1 | | 1.01| $\sqrt{3(1.01)-2} = \sqrt{3.03-2} = \sqrt{1.03} \approx 1.015$ | | 1.1 | $\sqrt{3.3-2} = \sqrt{1.3} \approx 1.14$ | Values approach 1 from both sides, confirming the limit is 1. **Final answer:** - $f(1) = 1$ - $\lim_{x \to 1} \sqrt{3x - 2} = 1$