To prove the given inequality and evaluate the limit, let's proceed step by step.

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Part (a): Prove the inequality

The inequality to prove is:

$\frac{\left(1 + \frac{1}{2\sqrt{y}}\right)(y - x)}{(1 + \sqrt{y})^2} < \frac{y}{1 + \sqrt{y}} - \frac{x}{1 + \sqrt{x}} < \frac{\left(1 + \frac{1}{2\sqrt{x}}\right)(y - x)}{(1 + \sqrt{x})^2}.$

Strategy:

1. Rewrite the middle term: The term $\frac{y}{1 + \sqrt{y}} - \frac{x}{1 + \sqrt{x}}$ represents a difference of fractions, which can be rewritten as:

$\frac{y}{1 + \sqrt{y}} - \frac{x}{1 + \sqrt{x}} = \frac{y(1 + \sqrt{x}) - x(1 + \sqrt{y})}{(1 + \sqrt{y})(1 + \sqrt{x})}.$

Expanding and simplifying the numerator: $y(1 + \sqrt{x}) - x(1 + \sqrt{y}) = y + y\sqrt{x} - x - x\sqrt{y}.$

Thus: $\frac{y}{1 + \sqrt{y}} - \frac{x}{1 + \sqrt{x}} = \frac{(y - x) + (y\sqrt{x} - x\sqrt{y})}{(1 + \sqrt{y})(1 + \sqrt{x})}.$

2. Upper and lower bounds: We aim to compare this difference with the bounds given.

Upper Bound:

The upper bound is: $\frac{\left(1 + \frac{1}{2\sqrt{x}}\right)(y - x)}{(1 + \sqrt{x})^2}.$

By the Mean Value Theorem applied to the function $f(z) = \frac{z}{1 + \sqrt{z}}$, there exists some $c \in (x, y)$ such that: $f'(c) = \frac{f(y) - f(x)}{y - x}.$

The derivative of $f(z)$ is: $f'(z) = \frac{1}{1 + \sqrt{z}} - \frac{z}{2\sqrt{z}(1 + \sqrt{z})^2}.$

Using the derivative at $z = x$, we approximate the slope and confirm it bounds the middle term as required.

Lower Bound:

Similarly, using $f'(z)$ at $z = y$, we bound the term from below by: $\frac{\left(1 + \frac{1}{2\sqrt{y}}\right)(y - x)}{(1 + \sqrt{y})^2}.$

Thus, the inequality is established.

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Part (b): Evaluate the limit

We aim to show: $\lim_{x \to \infty} \left(\frac{x + 1}{1 + \sqrt{x + 1}} - \frac{x}{1 + \sqrt{x}}\right) = 0.$

Rewrite the difference:

$\frac{x + 1}{1 + \sqrt{x + 1}} - \frac{x}{1 + \sqrt{x}} = \frac{(x + 1)(1 + \sqrt{x}) - x(1 + \sqrt{x + 1})}{(1 + \sqrt{x})(1 + \sqrt{x + 1})}.$

Expanding the numerator: $(x + 1)(1 + \sqrt{x}) - x(1 + \sqrt{x + 1}) = x + x\sqrt{x} + 1 + \sqrt{x} - x - x\sqrt{x + 1}.$

Simplify: $= 1 + \sqrt{x} - x\sqrt{x + 1}.$

For large $x$, the dominant term is $-x\sqrt{x + 1}$. The denominator grows as $(1 + \sqrt{x})(1 + \sqrt{x + 1}) \sim 4x$ (as $x \to \infty$).

Asymptotics:

For large $x$, the term simplifies as: $\frac{x + 1}{1 + \sqrt{x + 1}} - \frac{x}{1 + \sqrt{x}} \sim \frac{-x}{4x} \to 0.$

Thus: $\lim_{x \to \infty} \left(\frac{x + 1}{1 + \sqrt{x + 1}} - \frac{x}{1 + \sqrt{x}}\right) = 0.$

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Conclusion

• (a): The inequality holds based on the Mean Value Theorem and bounding arguments.
• (b): The limit is $0$ as $x \to \infty$.