1. **Problem statement:** We are given several limit problems and a function $f(x) = ax + b - \sqrt{x^2 + 1}$ to analyze. 2. **Inequality bounds:** For $m \leq \frac{1}{2 - \sin x} \leq M$, since $\sin x \in [-1,1]$, the denominator $2 - \sin x$ ranges from $1$ to $3$. Thus, $$\frac{1}{3} \leq \frac{1}{2 - \sin x} \leq 1,$$ so $m = \frac{1}{3}$ and $M = 1$. 3. **Functions:** - $f(x) = \frac{x}{1 - \sin x}$ - $g(x) = \frac{x + \sin x}{2 - \sin x}$ 4. **Limits:** **(a)** $\lim_{x \to 3} \frac{\sqrt{x^2 - 3}}{x - 3}$ - Direct substitution: numerator $= \sqrt{9 - 3} = \sqrt{6}$, denominator $= 0$, so form is $\frac{\sqrt{6}}{0}$, undefined. - Check if limit exists from left and right: - For $x > 3$, numerator real and positive, denominator positive small, limit $\to +\infty$. - For $x < 3$, numerator real but denominator negative small, limit $\to -\infty$. - So limit does not exist. **(b)** $\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$ - Factor numerator: $x^2 - 16 = (x - 4)(x + 4)$. - Simplify: $\frac{(x - 4)(x + 4)}{x - 4} = x + 4$ for $x \neq 4$. - Substitute $x = 4$: limit $= 8$. **(c)** $\lim_{x \to 8} \frac{\sqrt{2x - 4}}{\sqrt{x^2 + 1} - 3}$ - Substitute $x=8$: numerator $= \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$. - Denominator $= \sqrt{64 + 1} - 3 = \sqrt{65} - 3$. - Since denominator $\neq 0$, limit $= \frac{2\sqrt{3}}{\sqrt{65} - 3}$. **(d)** $\lim_{x \to 58} \frac{(2x + 5)^2 - 12x^2}{x - 58}$ - Expand numerator: $$(2x + 5)^2 - 12x^2 = (4x^2 + 20x + 25) - 12x^2 = -8x^2 + 20x + 25.$$ - Substitute $x=58$ numerator: $-8(58)^2 + 20(58) + 25 = -8(3364) + 1160 + 25 = -26912 + 1185 = -25727$. - Denominator $= 0$, so direct substitution gives $\frac{-25727}{0}$ undefined. - Use derivative (L'Hôpital's Rule): Numerator derivative: $-16x + 20$. Denominator derivative: $1$. - Evaluate at $x=58$: $-16(58) + 20 = -928 + 20 = -908$. - Limit $= -908$. **(e)** $\lim_{x \to \infty} \frac{x \sin x}{3x + \pi}$ - Denominator behaves like $3x$. - Numerator oscillates between $-x$ and $x$. - So fraction oscillates between $\frac{-x}{3x} = -\frac{1}{3}$ and $\frac{x}{3x} = \frac{1}{3}$. - Limit does not exist because of oscillation. **(f)** $\lim_{x \to -\infty} \frac{\sqrt{x^2 - 3}}{x - 3}$ - For large negative $x$, $\sqrt{x^2 - 3} \approx |x| = -x$ (since $x$ negative). - So numerator $\approx -x$. - Denominator $= x - 3 \approx x$. - So fraction $\approx \frac{-x}{x} = -1$. - Limit $= -1$. 5. **Function $f(x) = ax + b - \sqrt{x^2 + 1}$** **(1) Limit at $-\infty$ depending on $a$:** - For large negative $x$, $\sqrt{x^2 + 1} \approx |x| = -x$. - So $f(x) \approx ax + b - (-x) = (a + 1)x + b$. - If $a + 1 > 0$, then $f(x) \to -\infty$ as $x \to -\infty$. - If $a + 1 = 0$, then $f(x) \to b$ (finite limit). - If $a + 1 < 0$, then $f(x) \to +\infty$ as $x \to -\infty$. **(2) Find $a$ and $b$ so that line $2x - y + 2 = 0$ is asymptote at $-\infty$. - Rewrite line: $y = 2x + 2$. - For asymptote at $-\infty$, $f(x) - (2x + 2) \to 0$ as $x \to -\infty$. - Compute: $$f(x) - (2x + 2) = ax + b - \sqrt{x^2 + 1} - 2x - 2 = (a - 2)x + (b - 2) - \sqrt{x^2 + 1}.$$ - Approximate $\sqrt{x^2 + 1} = -x + \frac{1}{2|x|} + o(\frac{1}{x})$ as $x \to -\infty$. - So: $$f(x) - (2x + 2) \approx (a - 2)x + (b - 2) - (-x) = (a - 2 + 1)x + (b - 2) = (a - 1)x + (b - 2).$$ - For limit to be finite (0), coefficient of $x$ must be zero: $$a - 1 = 0 \Rightarrow a = 1.$$ - Then constant term must be zero: $$b - 2 = 0 \Rightarrow b = 2.$$ **Final answers:** - $m = \frac{1}{3}$, $M = 1$. - Limits: - (a) Does not exist. - (b) $8$. - (c) $\frac{2\sqrt{3}}{\sqrt{65} - 3}$. - (d) $-908$. - (e) Does not exist. - (f) $-1$. - For $f(x)$: - Limit at $-\infty$ depends on $a$ as above. - Asymptote line $2x - y + 2 = 0$ implies $a = 1$, $b = 2$.