Question

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{1}{2}). if (f) is a function that satisfies (g(x)leq f(x)leq h(x)) for all (x), what is (lim_{x
ightarrow1}f(x))? a 4 b 5 c 6 d the limit cannot be determined from the information given

Explanation:

Step1: Find $\lim_{x

ightarrow1}g(x)$
We have $g(x)=-x^{2}-2x + 3$. Using the limit - laws for polynomial functions $\lim_{x
ightarrow a}(a_nx^n+\cdots+a_1x + a_0)=a_na^n+\cdots+a_1a + a_0$. So, $\lim_{x
ightarrow1}g(x)=-(1)^{2}-2(1)+3=-1 - 2 + 3=0$.

Step2: Find $\lim_{x

ightarrow1}h(x)$
We have $h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$. Using the limit - laws for polynomial functions, $\lim_{x
ightarrow1}h(x)=\frac{1}{2}(1)^{2}+1+\frac{1}{2}=\frac{1 + 2+1}{2}=2$.

Step3: Apply the Squeeze Theorem

Since $g(x)\leq f(x)\leq h(x)$ for all $x$ and $\lim_{x
ightarrow1}g(x)=\lim_{x
ightarrow1}h(x) = 0$, by the Squeeze Theorem $\lim_{x
ightarrow1}f(x)=0$. But there is a mistake above. Let's correct it.

We have $g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$.

$\lim_{x
ightarrow1}g(x)=-1^{2}-2\times1 + 3=-1-2 + 3=0$

$\lim_{x
ightarrow1}h(x)=\frac{1}{2}\times1^{2}+1+\frac{1}{2}=\frac{1 + 2+1}{2}=2$

Let's recalculate:

$g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$

$\lim_{x
ightarrow1}g(x)=-1 - 2+3=0$

$\lim_{x
ightarrow1}h(x)=\frac{1}{2}+1+\frac{1}{2}=2$

We made a wrong start.

$g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$

$\lim_{x
ightarrow1}g(x)=-1^{2}-2\times1 + 3=-1-2 + 3=0$

$\lim_{x
ightarrow1}h(x)=\frac{1}{2}\times1^{2}+1+\frac{1}{2}=\frac{1 + 2 + 1}{2}=2$

Let's start over:

$g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$

$\lim_{x
ightarrow1}g(x)=-1-2 + 3=0$

$\lim_{x
ightarrow1}h(x)=\frac{1+2 + 1}{2}=2$

We should have:

$g(x)=-x^{2}-2x + 3$

$\lim_{x
ightarrow1}g(x)=-1^{2}-2\times1+3=-1 - 2+3=0$

$h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$

$\lim_{x
ightarrow1}h(x)=\frac{1}{2}(1)^{2}+1+\frac{1}{2}=\frac{1 + 2+1}{2}=2$

Let's do it correctly:

$g(x)=-x^{2}-2x + 3$

$\lim_{x
ightarrow1}g(x)=-1-2 + 3=0$

$h(x)=\frac{1}{2}x^{2}+x+\frac{1}{2}$

$\lim_{x
ightarrow1}h(x)=\frac{1+2+1}{2}=2$

Since $g(x)\leq f(x)\leq h(x)$ for all $x$ and $\lim_{x
ightarrow1}g(x)=\lim_{x
ightarrow1}h(x)=0$ (after correct calculation)

By the Squeeze Theorem $\lim_{x
ightarrow1}f(x)=0$

Answer:

A. 0