Question

the function g is given by ( g(x) = \frac{1}{x^2 - 4x + 5} ). the function h is given by ( h(x) = \frac{2x^2 - 8x + 10}{x^2 - 4x + 6} ). if f is a function that satisfies ( g(x) leq f(x) leq h(x) ) for ( 0 < x < 5 ), what is ( lim_{x \to 2} f(x) )?
options:
a 0
b 1
c 2
d the limit cannot be determined from the information given.

Explanation:

Step1: Find $\lim_{x \to 2} g(x)$

First, substitute $x = 2$ into the denominator of $g(x)=\frac{1}{x^{2}-4x + 5}$. The denominator is $2^{2}-4\times2 + 5=4 - 8 + 5 = 1$. So $\lim_{x \to 2} g(x)=\frac{1}{1}=1$? Wait, no, wait, let's recalculate the denominator: $x^{2}-4x + 5$ when $x = 2$: $2^2-4\times2 + 5=4 - 8 + 5 = 1$. So $g(2)=\frac{1}{1}=1$, and since $g(x)$ is a rational function and the denominator is not zero at $x = 2$, $\lim_{x \to 2} g(x)=g(2) = 1$? Wait, no, wait the problem is about $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 5}$. Wait, I misread $h(x)$'s denominator. Wait, $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 5}$? Wait, no, the original problem says $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 5}$? Wait, no, let's check again.

Wait, the function $g(x)=\frac{1}{x^{2}-4x + 5}$, and $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 5}$. Let's simplify $h(x)$: factor numerator and denominator.

Numerator of $h(x)$: $2x^{2}-8x + 10 = 2(x^{2}-4x + 5)$. Denominator of $h(x)$: $x^{2}-4x + 5$. So $h(x)=\frac{2(x^{2}-4x + 5)}{x^{2}-4x + 5}$ for $x^{2}-4x + 5
eq0$. Since $x^{2}-4x + 5=(x - 2)^{2}+1$, which is always positive (because square of real number is non - negative, plus 1 is positive), so $h(x)=2$ for all $x$ in the domain.

Now, $\lim_{x \to 2} g(x)$: substitute $x = 2$ into $g(x)=\frac{1}{x^{2}-4x + 5}$, we get $\frac{1}{(2)^{2}-4\times2 + 5}=\frac{1}{4 - 8 + 5}=\frac{1}{1}=1$.

$\lim_{x \to 2} h(x)$: since $h(x)=2$ for all $x$ (because the numerator and denominator cancel out, and the denominator is never zero), so $\lim_{x \to 2} h(x)=2$.

But wait, the Squeeze Theorem says that if $g(x)\leq f(x)\leq h(x)$ for all $x$ in an open interval containing $a$ (except possibly at $a$), and $\lim_{x \to a} g(x)=\lim_{x \to a} h(x)=L$, then $\lim_{x \to a} f(x)=L$. Wait, but here $\lim_{x \to 2} g(x)=1$ and $\lim_{x \to 2} h(x)=2$, which are not equal. Wait, no, I must have made a mistake. Wait, let's re - examine the functions.

Wait, the original problem: $g(x)=\frac{1}{x^{2}-4x + 5}$, $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 5}$. Let's compute $\lim_{x \to 2} g(x)$:

$x^{2}-4x + 5$ at $x = 2$: $4-8 + 5 = 1$, so $\lim_{x \to 2} g(x)=\frac{1}{1}=1$.

For $h(x)$: $\lim_{x \to 2}\frac{2x^{2}-8x + 10}{x^{2}-4x + 5}$. Substitute $x = 2$: numerator $2\times4-16 + 10=8 - 16 + 10 = 2$, denominator $4-8 + 5 = 1$, so $\lim_{x \to 2} h(x)=\frac{2}{1}=2$.

Wait, but the Squeeze Theorem requires that $\lim_{x \to a}g(x)=\lim_{x \to a}h(x)$. But here they are not equal. Wait, maybe I misread the functions. Wait, maybe the denominator of $h(x)$ is $x^{2}-4x + 6$? Wait, the original problem says: "The function $h$ is given by $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 6}$". Oh! I misread the denominator of $h(x)$. Let's correct that.

So $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 6}$. Now, let's compute $\lim_{x \to 2} g(x)$ and $\lim_{x \to 2} h(x)$.

First, $\lim_{x \to 2} g(x)$: $g(x)=\frac{1}{x^{2}-4x + 5}$. Substitute $x = 2$: $2^{2}-4\times2 + 5=4 - 8 + 5 = 1$, so $\lim_{x \to 2} g(x)=\frac{1}{1}=1$.

Now, $\lim_{x \to 2} h(x)$: substitute $x = 2$ into numerator and denominator of $h(x)$. Numerator: $2\times(2)^{2}-8\times2 + 10=8 - 16 + 10 = 2$. Denominator: $(2)^{2}-4\times2 + 6=4 - 8 + 6 = 2$. So $\lim_{x \to 2} h(x)=\frac{2}{2}=1$.

Ah! Now, by the Squeeze Theorem, since $g(x)\leq f(x)\leq h(x)$ for $0\lt x\lt5$, and $\lim_{x \to 2} g(x)=\lim_{x \to 2} h(x)=1$, then $\lim_{x \to 2} f(x)=1$. Wait, but the option B is 1. Wait, I think I made a mistake in the first reading of $h(x)$'s denominator. Let's confirm:

Original problem: "The function $h$ is given by $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 6}$". Yes, so denominator is $x^{2}-4x + 6$. Then:

$\lim_{x \to 2} h(x)=\frac{2(2)^2-8(2)+10}{(2)^2-4(2)+6}=\frac{8 - 16 + 10}{4 - 8 + 6}=\frac{2}{2}=1$.

$\lim_{x \to 2} g(x)=\frac{1}{(2)^2-4(2)+5}=\frac{1}{4 - 8 + 5}=1$.

So by Squeeze Theorem, since $g(x)\leq f(x)\leq h(x)$ for $0\lt x\lt5$, and $\lim_{x \to 2}g(x)=\lim_{x \to 2}h(x)=1$, then $\lim_{x \to 2}f(x)=1$. So the answer is B. 1.

Wait, I think my initial misreading of the denominator caused confusion. Let's redo the steps correctly:

Step1: Calculate $\lim_{x \to 2}g(x)$

For $g(x)=\frac{1}{x^{2}-4x + 5}$, substitute $x = 2$ into the denominator:
$x^{2}-4x + 5=2^{2}-4\times2 + 5=4 - 8 + 5 = 1$
So $\lim_{x \to 2}g(x)=\frac{1}{1}=1$

Step2: Calculate $\lim_{x \to 2}h(x)$

For $h(x)=\frac{2x^{2}-8x + 10}{x^{2}-4x + 6}$, substitute $x = 2$ into numerator and denominator:
Numerator: $2x^{2}-8x + 10=2\times2^{2}-8\times2 + 10=8 - 16 + 10 = 2$
Denominator: $x^{2}-4x + 6=2^{2}-4\times2 + 6=4 - 8 + 6 = 2$
So $\lim_{x \to 2}h(x)=\frac{2}{2}=1$

Step3: Apply the Squeeze Theorem

Since $g(x)\leq f(x)\leq h(x)$ for $0\lt x\lt5$, and $\lim_{x \to 2}g(x)=\lim_{x \to 2}h(x)=1$, by the Squeeze Theorem, $\lim_{x \to 2}f(x)=1$

Answer:

C. 2