we are given the following inequalities and are asked to find \\( \\lim _{x \\rightarrow 2} f(x) \\).\n\\( 2…

we are given the following inequalities and are asked to find \\( \\lim _{x \\rightarrow 2} f(x) \\).\n\\( 2 x-1 \\leq f(x) \\leq x^{2}-2 x+3 \\) for \\( x \\geq 0 \\)\nrecall the squeeze theorem.\nif \\( g(x) \\leq f(x) \\leq h(x) \\) when \\( x \\) is near \\( a \\) (except possibly at \\( a \\) ) and \\( \\lim _{x \\rightarrow a} g(x)=\\lim _{x \\rightarrow a} h(x)=l \\), then \\( \\lim _{x \\rightarrow a} f(x)=l \\).\nthat 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\nlet \\( 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.\n\\( \\lim _{x \\rightarrow 2}(g(x))=\\lim _{x \\rightarrow 2}(2 x-1) \\)\n\\( \\lim _{x \\rightarrow 2}(h(x))=\\lim _{x \\rightarrow 2}\\left(x^{2}-2 x+3\\right) \\)

we are given the following inequalities and are asked to find \\( \\lim _{x \\rightarrow 2} f(x) \\).\n\\( 2 x-1 \\leq f(x) \\leq x^{2}-2 x+3 \\) for \\( x \\geq 0 \\)\nrecall the squeeze theorem.\nif \\( g(x) \\leq f(x) \\leq h(x) \\) when \\( x \\) is near \\( a \\) (except possibly at \\( a \\) ) and \\( \\lim _{x \\rightarrow a} g(x)=\\lim _{x \\rightarrow a} h(x)=l \\), then \\( \\lim _{x \\rightarrow a} f(x)=l \\).\nthat 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\nlet \\( 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.\n\\( \\lim _{x \\rightarrow 2}(g(x))=\\lim _{x \\rightarrow 2}(2 x-1) \\)\n\\( \\lim _{x \\rightarrow 2}(h(x))=\\lim _{x \\rightarrow 2}\\left(x^{2}-2 x+3\\right) \\)

Answer

Explanation:

Step1: Calculate $\lim_{x\rightarrow2}g(x)$

Substitute $x = 2$ into $g(x)=2x - 1$. $$\lim_{x\rightarrow2}(2x - 1)=2\times2-1$$ $$=4 - 1=3$$

Step2: Calculate $\lim_{x\rightarrow2}h(x)$

Substitute $x = 2$ into $h(x)=x^{2}-2x + 3$. $$\lim_{x\rightarrow2}(x^{2}-2x + 3)=2^{2}-2\times2+3$$ $$=4-4 + 3=3$$

Since $g(x)\leq f(x)\leq h(x)$ for $x\geq0$ and $\lim_{x\rightarrow2}g(x)=\lim_{x\rightarrow2}h(x) = 3$, by the squeeze theorem.

Answer:

$3$