We are tasked with evaluating the following limit:

$\lim_{{x \to 1^+}} \frac{4x - 3}{\sqrt{x} - 1}$

Step 1: Direct Substitution

First, let's check what happens if we directly substitute $x = 1$:

$\frac{4(1) - 3}{\sqrt{1} - 1} = \frac{4 - 3}{1 - 1} = \frac{1}{0}$

This results in a division by zero, which suggests that the limit may approach infinity, but we need to determine if it's positive or negative infinity, or if it does not exist.

Step 2: Analyze the Behavior for $x \to 1^+$

As $x \to 1^+$, the numerator $4x - 3$ approaches $1$ from the positive side because $4x - 3 = 4(1^+) - 3 \approx 1$.

For the denominator $\sqrt{x} - 1$, we have $\sqrt{x}$ approaching 1, but for values slightly greater than 1, $\sqrt{x} > 1$. Therefore, $\sqrt{x} - 1$ will be positive and small.

Thus, for $x \to 1^+$, the numerator approaches 1 and the denominator approaches a small positive value. Therefore, the whole fraction grows large and positive.

Step 3: Conclusion

The limit tends towards positive infinity:

$\lim_{{x \to 1^+}} \frac{4x - 3}{\sqrt{x} - 1} = +\infty$

Would you like to see more details or have further questions about this solution?

Here are some related questions to explore:

1. What happens to the limit as $x \to 1^-$ instead?
2. How would you evaluate limits involving higher powers of $x$?
3. Can you apply L'Hopital's rule to this problem?
4. What would the limit be if the numerator were $4x^2 - 3$?
5. How does the concept of one-sided limits differ from general limits?

Tip: When you encounter division by zero in limits, analyze the signs and growth of the numerator and denominator near the point of interest to determine the limit's behavior.