The given limit expression is:

$\lim_{{h \to 0}} \frac{\sqrt{11 + h} - \sqrt{11}}{h}$

To solve this limit, we can multiply both the numerator and denominator by the conjugate of the numerator to simplify the expression. The conjugate of $\sqrt{11 + h} - \sqrt{11}$ is $\sqrt{11 + h} + \sqrt{11}$. Here's how it works:

Step-by-step solution:

1. Multiply the numerator and denominator by the conjugate:

$\frac{\sqrt{11 + h} - \sqrt{11}}{h} \cdot \frac{\sqrt{11 + h} + \sqrt{11}}{\sqrt{11 + h} + \sqrt{11}} = \frac{\left( \sqrt{11 + h} - \sqrt{11} \right) \left( \sqrt{11 + h} + \sqrt{11} \right)}{h \left( \sqrt{11 + h} + \sqrt{11} \right)}$

2. Apply the difference of squares to the numerator:

$\left( \sqrt{11 + h} \right)^2 - \left( \sqrt{11} \right)^2 = (11 + h) - 11 = h$

So, the expression becomes:

$\frac{h}{h \left( \sqrt{11 + h} + \sqrt{11} \right)}$

3. Simplify by canceling $h$ from the numerator and denominator:

$\frac{1}{\sqrt{11 + h} + \sqrt{11}}$

4. Now, take the limit as $h \to 0$:

$\lim_{{h \to 0}} \frac{1}{\sqrt{11 + h} + \sqrt{11}} = \frac{1}{\sqrt{11} + \sqrt{11}} = \frac{1}{2\sqrt{11}}$

Final Answer:

$\lim_{{h \to 0}} \frac{\sqrt{11 + h} - \sqrt{11}}{h} = \frac{1}{2\sqrt{11}}$

Let me know if you'd like further details or have any questions!

Here are some related questions you may find helpful:

1. How does rationalizing the numerator help in solving limits involving square roots?
2. What is the difference of squares, and why is it useful in this context?
3. How can we apply L'Hopital's Rule to this limit problem instead of conjugates?
4. How does the limit definition relate to the derivative of a function at a point?
5. What would the solution look like if the constant under the square root was different, such as 25 instead of 11?

Tip: Rationalizing expressions by multiplying with conjugates is a powerful tool to simplify square root limits!