Let's solve each part step by step.

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Part (a)

Prove that $\int_0^\infty f(x) \, dx$ exists if $\lim_{x \to \infty} \frac{f(x+1)}{f(x)}$ exists and is less than 1.

Proof:

1. Let $L = \lim_{x \to \infty} \frac{f(x+1)}{f(x)}$, where $L < 1$.

2. Since $L < 1$, choose a constant $c$ such that $L < c < 1$.

3. For sufficiently large $x$, say $x > X$, we have $\frac{f(x+1)}{f(x)} < c$.

4. By iteration, for $x > X$, $f(x+n) < c^n f(x).$

5. Consider the series $\sum_{n=0}^\infty f(x+n)$. Using the above inequality, this series is dominated by a geometric series: $f(x) \left( 1 + c + c^2 + \dots \right) = f(x) \cdot \frac{1}{1-c}.$ Hence, the tail of the series $\sum_{n=0}^\infty f(x+n)$ converges.

6. This implies that $f(x)$ decays fast enough for the integral $\int_0^\infty f(x) \, dx$ to converge.

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Part (b)

Prove that $\int_0^\infty f(x) \, dx$ exists if $\lim_{x \to \infty} \left( f(x) \right)^{\frac{1}{x}}$ exists and is less than 1.

Proof:

1. Let $L = \lim_{x \to \infty} \left( f(x) \right)^{\frac{1}{x}}$, where $L < 1$.
2. Since $L < 1$, choose $c$ such that $L < c < 1$.
3. For sufficiently large $x$, say $x > X$, we have $\left( f(x) \right)^{\frac{1}{x}} < c$. This implies $f(x) < c^x.$
4. Now consider the integral $\int_X^\infty f(x) \, dx$. Since $f(x)$ is dominated by $c^x$, and the integral of $c^x$ converges (as $c < 1$), the integral $\int_X^\infty f(x) \, dx$ also converges.
5. Adding the finite portion $\int_0^X f(x) \, dx$, we conclude that $\int_0^\infty f(x) \, dx$ converges.

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Part (c)

Prove that the integrals do not exist if the specified limits in (a) and (b) exist but exceed 1.

Case 1: Part (a)

1. Suppose $\lim_{x \to \infty} \frac{f(x+1)}{f(x)} = L$ with $L > 1$.
2. For sufficiently large $x$, $f(x+1) > L f(x)$. By iteration, $f(x+n) > L^n f(x).$
3. The series $\sum_{n=0}^\infty f(x+n)$ diverges because it is dominated by the divergent geometric series $\sum_{n=0}^\infty L^n$ (as $L > 1$).
4. Therefore, $\int_0^\infty f(x) \, dx$ does not converge.

Case 2: Part (b)

1. Suppose $\lim_{x \to \infty} \left( f(x) \right)^{\frac{1}{x}} = L$ with $L > 1$.
2. For sufficiently large $x$, $\left( f(x) \right)^{\frac{1}{x}} > L$. This implies $f(x) > L^x.$
3. The integral $\int_X^\infty f(x) \, dx$ diverges because $f(x)$ grows faster than any exponential function with base greater than 1.
4. Therefore, $\int_0^\infty f(x) \, dx$ does not converge.

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Summary:

1. $\int_0^\infty f(x) \, dx$ exists if:
   • $\lim_{x \to \infty} \frac{f(x+1)}{f(x)} < 1$, or
   • $\lim_{x \to \infty} \left( f(x) \right)^{\frac{1}{x}} < 1$.
2. The integral diverges if the corresponding limits exceed 1.

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Questions:

1. Why does $L < 1$ guarantee convergence for $\int_0^\infty f(x) \, dx$?
2. How does geometric series behavior help in analyzing $\int_0^\infty f(x) \, dx$?
3. What happens if $L = 1$ in both cases (a) and (b)?
4. Can we generalize this result for functions with different decay/growth rates?
5. How does the behavior of $f(x)$ at infinity influence the convergence of improper integrals?

Tip: Always compare $f(x)$ to a known function (e.g., exponential decay or growth) to estimate convergence of improper integrals