3.) (6pts) suppose that ( f ) is a function with domain ( (-infty, infty) ) and ( x^{2}+2 x-1 leq f(x) leq 6…

Explanation:

Step1: Find the limit of the lower - bound function

We have the lower - bound function (g(x)=x^{2}+2x - 1). To find (\lim_{x\rightarrow2}(x^{2}+2x - 1)), we use the limit rules (\lim_{x\rightarrow a}(u(x)+v(x)-w(x))=\lim_{x\rightarrow a}u(x)+\lim_{x\rightarrow a}v(x)-\lim_{x\rightarrow a}w(x)) and (\lim_{x\rightarrow a}x^{n}=a^{n}). (\lim_{x\rightarrow2}(x^{2}+2x - 1)=\lim_{x\rightarrow2}x^{2}+\lim_{x\rightarrow2}(2x)-\lim_{x\rightarrow2}1) Since (\lim_{x\rightarrow2}x^{2}=2^{2}), (\lim_{x\rightarrow2}(2x)=2\lim_{x\rightarrow2}x = 2\times2), and (\lim_{x\rightarrow2}1 = 1) (\lim_{x\rightarrow2}(x^{2}+2x - 1)=4 + 4-1=7)

Step2: Find the limit of the upper - bound function

We have the upper - bound function (h(x)=6x - 5). To find (\lim_{x\rightarrow2}(6x - 5)), we use the limit rules (\lim_{x\rightarrow a}(u(x)-v(x))=\lim_{x\rightarrow a}u(x)-\lim_{x\rightarrow a}v(x)) and (\lim_{x\rightarrow a}(cx)=c\lim_{x\rightarrow a}x) (where (c) is a constant). (\lim_{x\rightarrow2}(6x - 5)=\lim_{x\rightarrow2}(6x)-\lim_{x\rightarrow2}5) Since (\lim_{x\rightarrow2}(6x)=6\lim_{x\rightarrow2}x=6\times2) and (\lim_{x\rightarrow2}5 = 5) (\lim_{x\rightarrow2}(6x - 5)=12-5 = 7)

Step3: Apply the Squeeze Theorem

The Squeeze Theorem states that if (g(x)\leq f(x)\leq h(x)) for all (x) in some open interval containing (a) (except possibly at (x = a)) and (\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L), then (\lim_{x\rightarrow a}f(x)=L) Since (x^{2}+2x - 1\leq f(x)\leq6x - 5) for all (x\in(-\infty,\infty)) and (\lim_{x\rightarrow2}(x^{2}+2x - 1)=\lim_{x\rightarrow2}(6x - 5)=7)

Answer:

(\lim_{x\rightarrow2}f(x)=7)