please answer each question fully, providing all reasoning. you will be graded based on both mathematical…

please answer each question fully, providing all reasoning. you will be graded based on both mathematical correctness and clarity of writing. see the written portion rubric on d2l for more details.\nlet (b) denote the last non - zero digit of your ucid number. e.g. if your ucid is 9876543280 then (b = 8).\n1. let (f) be a function defined for all (x), and continuous at all values of (x) except at (-2) and (0). suppose that for all (x>10) we have that (bleq f(x)leq\frac{bx^{2}+x}{x^{2}-b}). moreover, suppose that the following is known:\n(lim_{x\rightarrow - 2^{-}}f(x)=3), (lim_{x\rightarrow - 2^{+}}f(x)=f(-2)=1), (lim_{x\rightarrow0^{-}}f(x)=f(0)=-b), (lim_{x\rightarrow0^{+}}f(x)=2).\n(a) does there always exist a value (c) on the interval ((-3,0)) such that (f(c)=0)? justify your answer.\n(b) does there always exist a value (c) on the interval ((-1,1)) such that (f(c)=0)? justify your answer.\n(c) evaluate the following limit:\n(lim_{x\rightarrowinfty}f(x)).\njustify your answer.\nhint: in (a) and (b) we want to know if there always exists such a value (c). finding a specific example where such a (c) exists does not show that something always happens. the only thing you are allowed to assume about (f(x)) is the information given in the prompt.

please answer each question fully, providing all reasoning. you will be graded based on both mathematical correctness and clarity of writing. see the written portion rubric on d2l for more details.\nlet (b) denote the last non - zero digit of your ucid number. e.g. if your ucid is 9876543280 then (b = 8).\n1. let (f) be a function defined for all (x), and continuous at all values of (x) except at (-2) and (0). suppose that for all (x>10) we have that (bleq f(x)leq\frac{bx^{2}+x}{x^{2}-b}). moreover, suppose that the following is known:\n(lim_{x\rightarrow - 2^{-}}f(x)=3), (lim_{x\rightarrow - 2^{+}}f(x)=f(-2)=1), (lim_{x\rightarrow0^{-}}f(x)=f(0)=-b), (lim_{x\rightarrow0^{+}}f(x)=2).\n(a) does there always exist a value (c) on the interval ((-3,0)) such that (f(c)=0)? justify your answer.\n(b) does there always exist a value (c) on the interval ((-1,1)) such that (f(c)=0)? justify your answer.\n(c) evaluate the following limit:\n(lim_{x\rightarrowinfty}f(x)).\njustify your answer.\nhint: in (a) and (b) we want to know if there always exists such a value (c). finding a specific example where such a (c) exists does not show that something always happens. the only thing you are allowed to assume about (f(x)) is the information given in the prompt.

Answer

Explanation:

Step1: Analyze part (a) - Intermediate Value Theorem

The Intermediate Value Theorem states that if a function (y = f(x)) is continuous on a closed - interval ([a,b]), and (k) is a number between (f(a)) and (f(b)), then there exists at least one number (c) in the interval ((a,b)) such that (f(c)=k). The function (f(x)) is continuous on ((-3, - 2)\cup(-2,0)). We know that (\lim_{x\rightarrow - 2^{-}}f(x)=3) and (\lim_{x\rightarrow - 2^{+}}f(x)=1) and (f(0)=-b). Since (f(x)) is not continuous on ([-3,0]) (discontinuous at (x = - 2)), we consider the fact that for (x>10), (b\leq f(x)\leq\frac{bx^{2}+x}{x^{2}-b}). (\lim_{x\rightarrow\infty}\frac{bx^{2}+x}{x^{2}-b}=\lim_{x\rightarrow\infty}\frac{b+\frac{1}{x}}{1 - \frac{b}{x^{2}}}=b). So for (x>10), (f(x)\geq b>0) (assuming (b>0)). We know that (f(x)) is discontinuous at (x=-2). Consider the left - hand limit (\lim_{x\rightarrow - 2^{-}}f(x) = 3>0) and the right - hand limit (\lim_{x\rightarrow - 2^{+}}f(x)=1>0) and (f(0)=-b<0). Since (f(x)) is discontinuous at (x = - 2), we cannot guarantee the existence of a (c\in(-3,0)) such that (f(c) = 0).

Step2: Analyze part (b) - Intermediate Value Theorem

The function (f(x)) is discontinuous at (x = 0). We know that (\lim_{x\rightarrow0^{-}}f(x)=-b<0) and (\lim_{x\rightarrow0^{+}}f(x)=2>0). But since (f(x)) is not continuous on ([-1,1]) (discontinuous at (x = 0)), we cannot use the Intermediate Value Theorem. So we cannot guarantee the existence of a (c\in(-1,1)) such that (f(c)=0).

Step3: Evaluate the limit in part (c)

For (x > 10), we have (b\leq f(x)\leq\frac{bx^{2}+x}{x^{2}-b}). We find (\lim_{x\rightarrow\infty}\frac{bx^{2}+x}{x^{2}-b}) using the rule for limits of rational functions. Divide both the numerator and denominator by (x^{2}): (\lim_{x\rightarrow\infty}\frac{bx^{2}+x}{x^{2}-b}=\lim_{x\rightarrow\infty}\frac{b+\frac{1}{x}}{1-\frac{b}{x^{2}}}=b). By the Squeeze Theorem, since (\lim_{x\rightarrow\infty}b = b) and (\lim_{x\rightarrow\infty}\frac{bx^{2}+x}{x^{2}-b}=b) and (b\leq f(x)\leq\frac{bx^{2}+x}{x^{2}-b}) for (x > 10), we have (\lim_{x\rightarrow\infty}f(x)=b).

Answer:

(a) No. The function is discontinuous at (x=-2) in the interval ((-3,0)) and we cannot apply the Intermediate Value Theorem. (b) No. The function is discontinuous at (x = 0) in the interval ((-1,1)) and we cannot apply the Intermediate Value Theorem. (c) (\lim_{x\rightarrow\infty}f(x)=b) by the Squeeze Theorem.