1. **State the problems:** We need to evaluate the following limits using a table of values: - $\lim_{x \to 1} (2x + 3)$ - $\lim_{x \to 2} (x^2 - 1)$ - $\lim_{x \to 3} \frac{10}{x - 2}$ - $\lim_{x \to 4} f(x)$ where $f(x) = \begin{cases} x + 3, & x \geq 4 \\ x - 1, & x < 4 \end{cases}$ 2. **Recall the limit definition and approach:** The limit $\lim_{x \to a} g(x)$ is the value that $g(x)$ approaches as $x$ gets arbitrarily close to $a$ from both sides. 3. **Construct tables of values near the points of interest:** --- **(a) For $\lim_{x \to 1} (2x + 3)$:** | $x$ | $2x + 3$ | |-----|----------| | 0.9 | $2(0.9)+3=4.8$ | | 0.99| $2(0.99)+3=4.98$ | | 1 | $2(1)+3=5$ | | 1.01| $2(1.01)+3=5.02$ | | 1.1 | $2(1.1)+3=5.2$ | As $x$ approaches 1, $2x+3$ approaches 5. --- **(b) For $\lim_{x \to 2} (x^2 - 1)$:** | $x$ | $x^2 - 1$ | |-----|-----------| | 1.9 | $1.9^2 - 1 = 3.61 - 1 = 2.61$ | | 1.99| $1.99^2 - 1 = 3.9601 - 1 = 2.9601$ | | 2 | $2^2 - 1 = 4 - 1 = 3$ | | 2.01| $2.01^2 - 1 = 4.0401 - 1 = 3.0401$ | | 2.1 | $2.1^2 - 1 = 4.41 - 1 = 3.41$ | As $x$ approaches 2, $x^2 - 1$ approaches 3. --- **(c) For $\lim_{x \to 3} \frac{10}{x - 2}$:** | $x$ | $\frac{10}{x - 2}$ | |-----|--------------------| | 2.9 | $\frac{10}{2.9 - 2} = \frac{10}{0.9} \approx 11.11$ | | 2.99| $\frac{10}{2.99 - 2} = \frac{10}{0.99} \approx 10.10$ | | 3 | $\frac{10}{3 - 2} = \frac{10}{1} = 10$ | | 3.01| $\frac{10}{3.01 - 2} = \frac{10}{1.01} \approx 9.90$ | | 3.1 | $\frac{10}{3.1 - 2} = \frac{10}{1.1} \approx 9.09$ | As $x$ approaches 3, $\frac{10}{x - 2}$ approaches 10. --- **(d) For $\lim_{x \to 4} f(x)$ where $f(x) = \begin{cases} x + 3, & x \geq 4 \\ x - 1, & x < 4 \end{cases}$:** | $x$ | $f(x)$ | |-----|---------| | 3.9 | $3.9 - 1 = 2.9$ | | 3.99| $3.99 - 1 = 2.99$ | | 4 | $4 + 3 = 7$ | | 4.01| $4.01 + 3 = 7.01$ | | 4.1 | $4.1 + 3 = 7.1$ | The left-hand limit as $x \to 4^-$ is $2.99$ (approaching 3), the right-hand limit as $x \to 4^+$ is $7$, so the two-sided limit does not exist because the left and right limits differ. 4. **Final answers:** - $\lim_{x \to 1} (2x + 3) = 5$ - $\lim_{x \to 2} (x^2 - 1) = 3$ - $\lim_{x \to 3} \frac{10}{x - 2} = 10$ - $\lim_{x \to 4} f(x)$ does not exist (left and right limits differ)