we have the following limits.\n\\( \\lim _ { x \\rightarrow 2 } ( 2 x - 1 ) = 3 \\) and \\( \\lim _ { x…

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

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

Answer

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\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x) = L), then (\lim_{x\rightarrow a}f(x)=L). Here, (g(x)=2x - 1), (h(x)=x^{2}-2x + 3), (a = 2), (\lim_{x\rightarrow 2}(2x - 1)=3) and (\lim_{x\rightarrow 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\rightarrow 2}(2x - 1)=\lim_{x\rightarrow 2}(x^{2}-2x + 3)=3), by the Squeeze Theorem, (\lim_{x\rightarrow 2}f(x)) must be equal to the common limit of the lower - bound and upper - bound functions.

Answer:

(3)