Show that
lim
n→∞ √
n + 1 −
√
n

= 0.

To show that

\[
\lim_{n \to \infty} \left( \sqrt{n+1} - \sqrt{n} \right) = 0,
\]

we can manipulate the expression in a way that makes it easier to evaluate as \( n \to \infty \).

### Step 1: Multiply by the Conjugate

The idea is to multiply and divide by the conjugate of \( \sqrt{n+1} - \sqrt{n} \), which will help simplify the expression. The conjugate of \( \sqrt{n+1} - \sqrt{n} \) is \( \sqrt{n+1} + \sqrt{n} \). Thus, we multiply and divide by \( \sqrt{n+1} + \sqrt{n} \):

\[
\sqrt{n+1} - \sqrt{n} = \frac{\left( \sqrt{n+1} - \sqrt{n} \right) \left( \sqrt{n+1} + \sqrt{n} \right)}{\sqrt{n+1} + \sqrt{n}}.
\]

### Step 2: Simplify the Numerator

Using the difference of squares identity \( (a - b)(a + b) = a^2 - b^2 \), we get:

\[
\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}.
\]

### Step 3: Evaluate the Limit

Now, we evaluate the limit of the simplified expression:

\[
\lim_{n \to \infty} \frac{1}{\sqrt{n+1} + \sqrt{n}}.
\]

As \( n \to \infty \), both \( \sqrt{n+1} \) and \( \sqrt{n} \) grow large. In particular, \( \sqrt{n+1} \approx \sqrt{n} \) when \( n \) is very large. So:

\[
\sqrt{n+1} + \sqrt{n} \approx 2\sqrt{n}.
\]

Thus, the expression becomes:

\[
\lim_{n \to \infty} \frac{1}{2\sqrt{n}} = 0.
\]

### Conclusion:

Therefore, we have shown that

\[
\lim_{n \to \infty} \left( \sqrt{n+1} - \sqrt{n} \right) = 0.
\]

---

Do you have any questions or would you like further details on any step?

Here are 5 related questions to deepen your understanding:

1. How does the difference of squares identity help simplify this limit?
2. What would happen if we tried to evaluate the limit without multiplying by the conjugate?
3. Can we apply this technique to limits involving other radical expressions?
4. How does this method relate to approximating functions for large values of \(n\)?
5. What does this limit tell us about the behavior of square roots for large numbers?

**Tip**: Multiplying by the conjugate is a common trick for simplifying limits and indeterminate forms involving radicals.

Show that lim n→∞ √(n + 1) − √(n) = 0.

This problem demonstrates how to evaluate the limit of √(n+1) − √(n) as n approaches infinity. Using algebraic manipulation, such as multiplying by the conjugate and applying the difference of squares, the solution shows that the limit evaluates to 0. Suitable for students in Grades 11-12 and early college.

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