1. The problem asks if Hayley's suggestion to use the functions \(g(x)=e^x\) and \(h(x)=e^{-x}\) to apply the squeeze theorem for function \(f(x)\) near \(x=0\) is correct. 2. The squeeze theorem requires that for values of \(x\) near a point (here \(0\)), one function \(h(x)\) is always less than or equal to \(f(x)\), and another function \(g(x)\) is always greater than or equal to \(f(x)\), and both \(g(x)\) and \(h(x)\) have the same limit at that point. 3. From the description, \(g(x)=e^x\) is increasing and passes through \((0,1)\), and \(h(x)=e^{-x}\) is decreasing and also passes through \((0,1)\). 4. The function \(f(x)\) has a U-shaped curve with an open circle at \((0,1)\), and values near \(0\) lie between \(g(x)\) and \(h(x)\). 5. Both \(g(x)\) and \(h(x)\) approach \(1\) as \(x \to 0\), since \(e^0=1\) and \(e^{-0}=1\). 6. Since \(h(x) \leq f(x) \leq g(x)\) near \(0\) and \(\lim_{x \to 0} g(x) = \lim_{x \to 0} h(x) = 1\), the squeeze theorem applies and \(\lim_{x \to 0} f(x) = 1\). 7. Therefore, Hayley's suggestion is correct. **Final answer:** A Yes, Hayley's suggestion seems to be correct.