Evaluate the limit: lim (n to infinity) sqrt(n) / (sqrt(n+1) - sqrt(n))

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{n+1} + \sqrt{n}$.

$$\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1} - \sqrt{n}} \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$$

The denominator uses the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

$$(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n}) = (\sqrt{n+1})^2 - (\sqrt{n})^2 = (n+1) - n = 1$$

The numerator becomes:

$$\sqrt{n}(\sqrt{n+1} + \sqrt{n}) = \sqrt{n(n+1)} + \sqrt{n^2} = \sqrt{n^2+n} + n$$

So the expression simplifies to:

$$\lim_{n \to \infty} (\sqrt{n^2+n} + n)$$

As $n \to \infty$, both terms in the expression approach infinity:

$$\lim_{n \to \infty} \sqrt{n^2+n} = \infty$$ $$\lim_{n \to \infty} n = \infty$$

Therefore, their sum also approaches infinity:

$$\lim_{n \to \infty} (\sqrt{n^2+n} + n) = \infty + \infty = \infty$$

The limit evaluates to $\infty$.

The final answer is $\text{(D) } \infty$.