Question

we are given the following inequalities and are asked to find \\( \lim _{x \
ightarrow 2} f(x) \\).
\\( 2 x-1 \leq f(x) \leq x^{2}-2 x+3 \\) for \\( x \geq 0 \\)
recall the squeeze theorem.
if \\( g(x) \leq f(x) \leq h(x) \\) when \\( x \\) is near \\( a \\) (except possibly at \\( a \\) ) and \\( \lim _{x \
ightarrow a} g(x)=\lim _{x \
ightarrow a} h(x)=l \\), then \\( \lim _{x \
ightarrow a} f(x)=l \\).
that is, \\( f \\) must be greater than or equal the limit \\( l \\) because \\( f \geq g \\) and also \\( f \\) must be less than or equal to the limit \\( l \\) because \\( f \leq h \\). if these are simultaneously true, it muel
let \\( g(x)=2 x-1 \\) and \\( h(x)=x^{2}-2 x+3 \\). as these functions are both polynomials, the following limits may be evaluated by direct substitution.
\\( \lim _{x \
ightarrow 2}(g(x))=\lim _{x \
ightarrow 2}(2 x-1) \\)
\\( \lim _{x \
ightarrow 2}(h(x))=\lim _{x \
ightarrow 2}\left(x^{2}-2 x+3\
ight) \\)

Explanation:

Step1: Calculate $\lim_{x

ightarrow2}g(x)$
Substitute $x = 2$ into $g(x)=2x - 1$.

$$\lim_{x ightarrow2}(2x - 1)=2\times2-1$$

$$=4 - 1=3$$

Step2: Calculate $\lim_{x

ightarrow2}h(x)$
Substitute $x = 2$ into $h(x)=x^{2}-2x + 3$.

$$\lim_{x ightarrow2}(x^{2}-2x + 3)=2^{2}-2\times2+3$$

$$=4-4 + 3=3$$

Since $g(x)\leq f(x)\leq h(x)$ for $x\geq0$ and $\lim_{x
ightarrow2}g(x)=\lim_{x
ightarrow2}h(x) = 3$, by the squeeze theorem.

Answer:

$3$