We are tasked with finding the following limit:

$\lim_{h \to 0^+} \frac{\sqrt{h^2 + 4h + 3} - \sqrt{3}}{h}$

Step-by-Step Solution:

Step 1: Recognize the Indeterminate Form

As $h \to 0^+$, directly substituting $h = 0$ in the expression gives:

$\frac{\sqrt{0^2 + 4(0) + 3} - \sqrt{3}}{0} = \frac{\sqrt{3} - \sqrt{3}}{0} = \frac{0}{0}$

This is an indeterminate form, which suggests that we need to manipulate the expression further.

Step 2: Multiply by the Conjugate

To resolve the indeterminate form, multiply both the numerator and denominator by the conjugate of the numerator. The conjugate of $\sqrt{h^2 + 4h + 3} - \sqrt{3}$ is $\sqrt{h^2 + 4h + 3} + \sqrt{3}$.

Thus, multiply the expression by:

$\frac{\sqrt{h^2 + 4h + 3} + \sqrt{3}}{\sqrt{h^2 + 4h + 3} + \sqrt{3}}$

This gives:

$\lim_{h \to 0^+} \frac{\left( \sqrt{h^2 + 4h + 3} - \sqrt{3} \right)\left( \sqrt{h^2 + 4h + 3} + \sqrt{3} \right)}{h \left( \sqrt{h^2 + 4h + 3} + \sqrt{3} \right)}$

Step 3: Simplify the Numerator

The numerator simplifies as follows, using the difference of squares formula $(a - b)(a + b) = a^2 - b^2$:

$\left( \sqrt{h^2 + 4h + 3} \right)^2 - \left( \sqrt{3} \right)^2 = (h^2 + 4h + 3) - 3 = h^2 + 4h$

So the expression becomes:

$\lim_{h \to 0^+} \frac{h^2 + 4h}{h \left( \sqrt{h^2 + 4h + 3} + \sqrt{3} \right)}$

Step 4: Factor and Simplify

Factor the numerator:

$h^2 + 4h = h(h + 4)$

Now the expression simplifies to:

$\lim_{h \to 0^+} \frac{h(h + 4)}{h \left( \sqrt{h^2 + 4h + 3} + \sqrt{3} \right)}$

Cancel the $h$ terms:

$\lim_{h \to 0^+} \frac{h + 4}{\sqrt{h^2 + 4h + 3} + \sqrt{3}}$

Step 5: Evaluate the Limit

As $h \to 0^+$, substitute $h = 0$ into the remaining expression:

$\frac{0 + 4}{\sqrt{0^2 + 4(0) + 3} + \sqrt{3}} = \frac{4}{\sqrt{3} + \sqrt{3}} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$

To rationalize the denominator, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$:

$\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

Thus, the final value of the limit is:

$\boxed{\frac{2\sqrt{3}}{3}}$

Let me know if you'd like further details or clarifications!

Follow-up Questions:

1. How does multiplying by the conjugate help in resolving indeterminate forms?
2. Why does the indeterminate form $\frac{0}{0}$ occur frequently in limits involving square roots?
3. What other techniques, besides multiplying by the conjugate, can be used to resolve limits with square roots?
4. How would the solution differ if the limit were approached as $h \to 0^-$?
5. Can you explain the importance of rationalizing the denominator in the final step?

Tip:

When faced with limits involving square roots, multiplying by the conjugate is often a powerful tool to simplify the expression and resolve indeterminate forms.