Question

3.) (6pts) suppose that ( f ) is a function with domain ( (-infty, infty) ) and ( x^{2}+2 x-1 leq f(x) leq 6 x-5 ) for all ( x in(-infty, infty) ). use the squeeze theorem to evaluate the following limit. ( lim _{x
ightarrow 2} f(x) )

Explanation:

Step1: Find the limit of the lower - bound function

We have the lower - bound function \(g(x)=x^{2}+2x - 1\).
To find \(\lim_{x
ightarrow2}(x^{2}+2x - 1)\), we use the limit rules \(\lim_{x
ightarrow a}(u(x)+v(x)-w(x))=\lim_{x
ightarrow a}u(x)+\lim_{x
ightarrow a}v(x)-\lim_{x
ightarrow a}w(x)\) and \(\lim_{x
ightarrow a}x^{n}=a^{n}\).
\(\lim_{x
ightarrow2}(x^{2}+2x - 1)=\lim_{x
ightarrow2}x^{2}+\lim_{x
ightarrow2}(2x)-\lim_{x
ightarrow2}1\)
Since \(\lim_{x
ightarrow2}x^{2}=2^{2}\), \(\lim_{x
ightarrow2}(2x)=2\lim_{x
ightarrow2}x = 2\times2\), and \(\lim_{x
ightarrow2}1 = 1\)
\(\lim_{x
ightarrow2}(x^{2}+2x - 1)=4 + 4-1=7\)

Step2: Find the limit of the upper - bound function

We have the upper - bound function \(h(x)=6x - 5\).
To find \(\lim_{x
ightarrow2}(6x - 5)\), we use the limit rules \(\lim_{x
ightarrow a}(u(x)-v(x))=\lim_{x
ightarrow a}u(x)-\lim_{x
ightarrow a}v(x)\) and \(\lim_{x
ightarrow a}(cx)=c\lim_{x
ightarrow a}x\) (where \(c\) is a constant).
\(\lim_{x
ightarrow2}(6x - 5)=\lim_{x
ightarrow2}(6x)-\lim_{x
ightarrow2}5\)
Since \(\lim_{x
ightarrow2}(6x)=6\lim_{x
ightarrow2}x=6\times2\) and \(\lim_{x
ightarrow2}5 = 5\)
\(\lim_{x
ightarrow2}(6x - 5)=12-5 = 7\)

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 \(\lim_{x
ightarrow a}g(x)=\lim_{x
ightarrow a}h(x)=L\), then \(\lim_{x
ightarrow a}f(x)=L\)
Since \(x^{2}+2x - 1\leq f(x)\leq6x - 5\) for all \(x\in(-\infty,\infty)\) and \(\lim_{x
ightarrow2}(x^{2}+2x - 1)=\lim_{x
ightarrow2}(6x - 5)=7\)

Answer:

\(\lim_{x
ightarrow2}f(x)=7\)