To solve the limit

$\lim_{{t \to 4}} \frac{{t - \sqrt{3t + 4}}}{{4 - t}}$

we can start by simplifying the expression. A good approach here is to multiply the numerator and denominator by the conjugate of the numerator to eliminate the square root. This will help us deal with the indeterminate form.

Step-by-Step Solution

1. Multiply by the Conjugate: Multiply both the numerator and the denominator by the conjugate of the numerator: $(t + \sqrt{3t + 4})$.

   $\lim_{{t \to 4}} \frac{{t - \sqrt{3t + 4}}}{{4 - t}} \cdot \frac{{t + \sqrt{3t + 4}}}{{t + \sqrt{3t + 4}}}$

2. Simplify the Numerator: The numerator becomes a difference of squares:

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

   Simplifying further:

   $t^2 - 3t - 4$

   Factor $t^2 - 3t - 4$:

   $t^2 - 3t - 4 = (t - 4)(t + 1)$

3. Rewrite the Limit: Substitute back into the limit:

   $\lim_{{t \to 4}} \frac{{(t - 4)(t + 1)}}{{(4 - t)(t + \sqrt{3t + 4})}}$

4. Simplify the Fraction: Notice that $4 - t = -(t - 4)$, so we can cancel $(t - 4)$ terms:

   $\lim_{{t \to 4}} \frac{{(t + 1)}}{{-(t + \sqrt{3t + 4})}}$

5. Evaluate the Limit: Substitute $t = 4$:

   $= \frac{{4 + 1}}{{-(4 + \sqrt{3 \cdot 4 + 4})}} = \frac{5}{-(4 + \sqrt{12 + 4})} = \frac{5}{-(4 + \sqrt{16})} = \frac{5}{-(4 + 4)} = \frac{5}{-8} = -\frac{5}{8}$

Final Answer:

$\lim_{{t \to 4}} \frac{{t - \sqrt{3t + 4}}}{{4 - t}} = -\frac{5}{8}$

Would you like a detailed explanation of any of these steps or have any questions?

Here are some related questions for further exploration:

1. How do conjugates help in simplifying expressions involving square roots?
2. Why is factoring useful in limit problems involving indeterminate forms?
3. Can you use L'Hôpital's Rule to solve this limit? If so, how would it work here?
4. What are the common techniques to simplify limits that result in an indeterminate form?
5. How does the limit change if we modify the function slightly, such as changing the numerator?

Tip: When dealing with square roots in limits, always consider multiplying by the conjugate, as it can simplify the problem by removing the root.