The problem you've provided involves the limit of the expression:

$\lim_{{x \to 1}} \frac{{\sqrt{x}}}{{x + 1}}$

We can evaluate this limit by directly substituting $x = 1$, as no indeterminate form (like $\frac{0}{0}$) arises. Let's compute it:

Substituting $x = 1$ into the expression:

$\frac{{\sqrt{1}}}{{1 + 1}} = \frac{1}{2}$

So, the limit is:

$\boxed{\frac{1}{2}}$

Would you like more details or have any questions?

Here are some related questions to explore further:

1. How would the result change if the denominator were $x - 1$?
2. What is the general approach to handling limits involving square roots?
3. How does L'Hôpital's Rule apply if this were an indeterminate form?
4. How do absolute values in limits affect the solution process?
5. Can limits with radical expressions lead to continuity issues?

Tip: Always check for indeterminate forms (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$) when evaluating limits, as these require special techniques like factoring, L'Hôpital's Rule, or algebraic manipulation.