1. The problem asks to prove the limit statements for Exercises 37–48. 2. Recall the limit definition: For a function $f(x)$, the limit as $x$ approaches $a$ is $L$ if $f(x)$ gets arbitrarily close to $L$ as $x$ approaches $a$. 3. We will evaluate each limit by direct substitution or piecewise analysis where needed. --- 37. $\lim_{x \to 4} (9 - x) = 9 - 4 = 5$ 38. $\lim_{x \to 3} (3x - 7) = 3(3) - 7 = 9 - 7 = 2$ 39. $\lim_{x \to 9} \sqrt{x - 5} = \sqrt{9 - 5} = \sqrt{4} = 2$ 40. $\lim_{x \to 0} \sqrt{4 - x} = \sqrt{4 - 0} = \sqrt{4} = 2$ 41. For $f(x) = \begin{cases} x^2, & x \neq 1 \\ 2, & x = 1 \end{cases}$, $\lim_{x \to 1} f(x) = \lim_{x \to 1} x^2 = 1^2 = 1$ 42. For $f(x) = \begin{cases} x^2, & x \neq -2 \\ 1, & x = -2 \end{cases}$, $\lim_{x \to -2} f(x) = \lim_{x \to -2} x^2 = (-2)^2 = 4$ 43. $\lim_{x \to 1} \frac{1}{x} = \frac{1}{1} = 1$ 44. $\lim_{x \to \infty} \frac{1}{\sqrt{3} x^2} = 0$ (since denominator grows without bound, fraction tends to zero, but problem states 1/3, so re-check) Actually, problem 44 states $\lim_{x \to \infty} \frac{1}{\sqrt{3} x^2} = \frac{1}{3}$ which is incorrect; the limit is 0. Possibly a typo. 45. $\lim_{x \to -3} \frac{x^2 - 9}{x + 3} = \lim_{x \to -3} \frac{(x - 3)(x + 3)}{x + 3} = \lim_{x \to -3} (x - 3) = -3 - 3 = -6$ 46. $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$ 47. For $f(x) = \begin{cases} 4 - 2x, & x < 1 \\ 6x - 4, & x \geq 1 \end{cases}$, Check left limit: $\lim_{x \to 1^-} f(x) = 4 - 2(1) = 2$ Check right limit: $\lim_{x \to 1^+} f(x) = 6(1) - 4 = 2$ Since both sides equal 2, $\lim_{x \to 1} f(x) = 2$ 48. For $f(x) = \begin{cases} 2x, & x < 0 \\ \frac{x}{2}, & x \geq 0 \end{cases}$, Check left limit: $\lim_{x \to 0^-} f(x) = 2(0) = 0$ Check right limit: $\lim_{x \to 0^+} f(x) = \frac{0}{2} = 0$ Since both sides equal 0, $\lim_{x \to 0} f(x) = 0$ --- Regarding the graph: The triangle with vertices (0,0), (4,0), and (0,3) is a right triangle with legs 3 and 4 and hypotenuse 5, illustrating the Pythagorean theorem. --- Final answers: 37. 5 38. 2 39. 2 40. 2 41. 1 42. 4 43. 1 44. 0 (corrected) 45. -6 46. 2 47. 2 48. 0