let $g$ and $h$ be the functions defined by $g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{13}{2}$…

let $g$ and $h$ be the functions defined by $g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{13}{2}$. if $f$ is a function that satisfies $g(x)leq f(x)leq h(x)$ for all $x$, what is $lim_{x\rightarrow - 1}f(x)$?\na 4\nb 5\nc 6\nd the limit cannot be determined from the information given.

let $g$ and $h$ be the functions defined by $g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{13}{2}$. if $f$ is a function that satisfies $g(x)leq f(x)leq h(x)$ for all $x$, what is $lim_{x\rightarrow - 1}f(x)$?\na 4\nb 5\nc 6\nd the limit cannot be determined from the information given.

Answer

Explanation:

Step1: Recall the Squeeze - Theorem

The Squeeze - Theorem states that if (g(x)\leq f(x)\leq h(x)) for all (x) in an 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). Here (a=- 1).

Step2: Calculate (\lim_{x\rightarrow - 1}g(x))

Substitute (x=-1) into (g(x)=-x^{2}-2x + 3). [ \begin{align*} \lim_{x\rightarrow - 1}g(x)&=-(-1)^{2}-2\times(-1)+3\ &=-1 + 2+3\ &=4 \end{align*} ]

Step3: Calculate (\lim_{x\rightarrow - 1}h(x))

Substitute (x = - 1) into (h(x)=\frac{1}{2}x^{2}+x+\frac{13}{2}). [ \begin{align*} \lim_{x\rightarrow - 1}h(x)&=\frac{1}{2}\times(-1)^{2}+(-1)+\frac{13}{2}\ &=\frac{1}{2}-1+\frac{13}{2}\ &=\frac{1 - 2+13}{2}\ &=\frac{12}{2}\ &=4 \end{align*} ]

Step4: Apply the Squeeze - Theorem

Since (g(x)\leq f(x)\leq h(x)) for all (x) and (\lim_{x\rightarrow - 1}g(x)=\lim_{x\rightarrow - 1}h(x)=4), by the Squeeze - Theorem, (\lim_{x\rightarrow - 1}f(x)=4).

Answer:

A. 4