Question

we have the following limits.
\\( \lim _ { x \
ightarrow 2 } ( 2 x - 1 ) = 3 \\) and \\( \lim _ { x \
ightarrow 2 } \left( x ^ { 2 } - 2 x + 3 \
ight) = 3 \\)
we are also given the following.
\\( 2 x - 1 \leq f ( x ) \leq x ^ { 2 } - 2 x + 3 \\) for \\( x \geq 0 \\)
to conclude, utilize the squeeze theorem to find \\( \lim _ { x \
ightarrow 2 } f ( x ) \\).

Explanation:

Step1: Recall the Squeeze Theorem

If \(g(x)\leq f(x)\leq h(x)\) for all \(x\) in some open interval containing \(a\) (except possibly at \(a\) itself), and \(\lim_{x
ightarrow a}g(x)=\lim_{x
ightarrow a}h(x) = L\), then \(\lim_{x
ightarrow a}f(x)=L\).
Here, \(g(x)=2x - 1\), \(h(x)=x^{2}-2x + 3\), \(a = 2\), \(\lim_{x
ightarrow 2}(2x - 1)=3\) and \(\lim_{x
ightarrow 2}(x^{2}-2x + 3)=3\).

Step2: Apply the Squeeze Theorem

Since \(2x-1\leq f(x)\leq x^{2}-2x + 3\) for \(x\geq0\) (and in particular in an open interval around \(x = 2\)) and \(\lim_{x
ightarrow 2}(2x - 1)=\lim_{x
ightarrow 2}(x^{2}-2x + 3)=3\), by the Squeeze Theorem, \(\lim_{x
ightarrow 2}f(x)\) must be equal to the common limit of the lower - bound and upper - bound functions.

Answer:

\(3\)