We are tasked with solving the following limit:

$\lim_{h \to 0} \frac{\sqrt{h^2 + 13h + 10} - \sqrt{10}}{h}.$

Step-by-Step Solution:

1. Multiply and Divide by the Conjugate: To eliminate the square roots in the numerator, we multiply and divide by the conjugate of the numerator:

   $\frac{\sqrt{h^2 + 13h + 10} - \sqrt{10}}{h} \cdot \frac{\sqrt{h^2 + 13h + 10} + \sqrt{10}}{\sqrt{h^2 + 13h + 10} + \sqrt{10}}.$

   This simplifies the numerator using the difference of squares:

   $\frac{\left(\sqrt{h^2 + 13h + 10}\right)^2 - \left(\sqrt{10}\right)^2}{h \cdot \left(\sqrt{h^2 + 13h + 10} + \sqrt{10}\right)}.$

2. Simplify the Numerator: Expand the squares in the numerator:

   $\left(\sqrt{h^2 + 13h + 10}\right)^2 = h^2 + 13h + 10, \quad \left(\sqrt{10}\right)^2 = 10.$

   Subtracting these gives:

   $h^2 + 13h + 10 - 10 = h^2 + 13h.$

   So the fraction becomes:

   $\frac{h^2 + 13h}{h \cdot \left(\sqrt{h^2 + 13h + 10} + \sqrt{10}\right)}.$

3. Factor and Cancel $h$ in the Numerator: Factor $h$ out of the numerator:

   $h^2 + 13h = h(h + 13).$

   This allows cancellation of $h$ from the numerator and denominator (valid as $h \neq 0$):

   $\frac{h + 13}{\sqrt{h^2 + 13h + 10} + \sqrt{10}}.$

4. Evaluate the Limit as $h \to 0$: As $h \to 0$, the numerator $h + 13$ approaches $13$, and the square root in the denominator simplifies as follows:

   $\sqrt{h^2 + 13h + 10} \to \sqrt{10}.$

   Thus, the denominator becomes:

   $\sqrt{h^2 + 13h + 10} + \sqrt{10} \to \sqrt{10} + \sqrt{10} = 2\sqrt{10}.$

   Therefore, the limit is:

   $\lim_{h \to 0} \frac{h + 13}{\sqrt{h^2 + 13h + 10} + \sqrt{10}} = \frac{13}{2\sqrt{10}}.$

5. Simplify the Final Expression: Rationalizing the denominator is optional, but the answer matches choice II in the provided options:

   $\boxed{\frac{13}{2\sqrt{10}}}.$

Would you like more details or have any questions?

Related Questions:

1. How do you multiply and divide by a conjugate in limits?
2. What are the properties of square roots that simplify limits?
3. How do you determine the dominant term in a binomial expansion near zero?
4. How can you rationalize the denominator in expressions involving square roots?
5. Why is canceling $h$ valid when $h \neq 0$?

Tip:

When dealing with limits involving square roots, multiplying by the conjugate is a powerful technique to simplify the expression.