Questions: Find the limit of f(x) as x approaches infinity if, for all x>1, (10e^x - 23) / (2e^x) < f(x) < 5√x / √(x-1)

Solution Steps

To find the limit of \( f(x) \) as \( x \) approaches infinity, we need to analyze the behavior of the given bounds for \( f(x) \). Specifically, we will evaluate the limits of the lower bound \(\frac{10 e^{x}-23}{2 e^{x}}\) and the upper bound \(\frac{5 \sqrt{x}}{\sqrt{x-1}}\) as \( x \) approaches infinity. If both limits converge to the same value, then \( \lim_{x \to \infty} f(x) \) will be that value.

1. Evaluate the limit of the lower bound \(\frac{10 e^{x}-23}{2 e^{x}}\) as \( x \) approaches infinity.
2. Evaluate the limit of the upper bound \(\frac{5 \sqrt{x}}{\sqrt{x-1}}\) as \( x \) approaches infinity.
3. Compare the two limits to determine the limit of \( f(x) \).

We start by evaluating the limit of the lower bound as \( x \) approaches infinity: \[ \lim_{x \to \infty} \frac{10 e^{x} - 23}{2 e^{x}} \] Simplifying the expression inside the limit: \[ \frac{10 e^{x} - 23}{2 e^{x}} = \frac{10 e^{x}}{2 e^{x}} - \frac{23}{2 e^{x}} = 5 - \frac{23}{2 e^{x}} \] As \( x \) approaches infinity, \( \frac{23}{2 e^{x}} \) approaches 0. Therefore: \[ \lim_{x \to \infty} \left( 5 - \frac{23}{2 e^{x}} \right) = 5 \]

Next, we evaluate the limit of the upper bound as \( x \) approaches infinity: \[ \lim_{x \to \infty} \frac{5 \sqrt{x}}{\sqrt{x-1}} \] Simplifying the expression inside the limit: \[ \frac{5 \sqrt{x}}{\sqrt{x-1}} = 5 \cdot \frac{\sqrt{x}}{\sqrt{x-1}} = 5 \cdot \sqrt{\frac{x}{x-1}} \] As \( x \) approaches infinity, \( \frac{x}{x-1} \) approaches 1. Therefore: \[ \lim_{x \to \infty} 5 \cdot \sqrt{\frac{x}{x-1}} = 5 \cdot \sqrt{1} = 5 \]

Since both the lower and upper bounds converge to the same value as \( x \) approaches infinity: \[ \lim_{x \to \infty} \frac{10 e^{x} - 23}{2 e^{x}} = 5 \quad \text{and} \quad \lim_{x \to \infty} \frac{5 \sqrt{x}}{\sqrt{x-1}} = 5 \] We conclude that: \[ \lim_{x \to \infty} f(x) = 5 \]

Final Answer