110. Does f grow faster, slower, or at the same rate as g as x→∞? Give reasons for your answers.
f. f(x) = sech(x), g(x) = e^(-x)

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Recall the definitions of the functions: \(f(x) = \text{sech}(x)\) and \(g(x) = e^{-x}\). The hyperbolic secant function is defined as \(\text{sech}(x) = \frac{2}{e^{x} + e^{-x}}\).

Analyze the behavior of \(f(x)\) as \(x \to \infty\). Since \(e^{x}\) grows very large, the term \(e^{x} + e^{-x}\) is dominated by \(e^{x}\), so \(\text{sech}(x) \approx \frac{2}{e^{x}} = 2e^{-x}\) for large \(x\).

Compare this approximation of \(f(x)\) to \(g(x) = e^{-x}\). Notice that \(f(x)\) behaves like \(2e^{-x}\), which is just a constant multiple of \(g(x)\) as \(x \to \infty\).

Since \(f(x)\) and \(g(x)\) differ only by a constant factor in their dominant terms for large \(x\), they decay at the same exponential rate as \(x \to \infty\).

Conclude that \(f\) and \(g\) grow (or decay) at the same rate as \(x \to \infty\) because their leading terms have the same exponential order.

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Asymptotic behavior describes how functions behave as the input approaches a limit, often infinity. Understanding this helps compare growth rates by analyzing limits of ratios or differences as x→∞, revealing which function dominates or if they grow at similar rates.

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The hyperbolic secant, sech(x), is defined as 1/cosh(x), where cosh(x) = (e^x + e^{-x})/2. As x→∞, cosh(x) grows exponentially, so sech(x) approaches zero exponentially, behaving similarly to e^{-x} but with a specific constant factor.

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Exponential decay functions like e^{-x} decrease rapidly as x→∞. Comparing f(x) = sech(x) to g(x) = e^{-x} involves examining their rates of decay by considering limits of their ratio, which determines if one decays faster, slower, or at the same rate.

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