We are tasked with evaluating the following limit:

$\lim_{{x \to 1}} \frac{\left(\frac{1}{x+1} - \frac{1}{2}\right)}{x-1}$

Step 1: Simplify the expression inside the limit

The numerator is $\frac{1}{x+1} - \frac{1}{2}$, and we need to combine the two terms into a single fraction. Using a common denominator, we get:

$\frac{1}{x+1} - \frac{1}{2} = \frac{2 - (x+1)}{2(x+1)} = \frac{2 - x - 1}{2(x+1)} = \frac{1 - x}{2(x+1)}$

Thus, the original limit becomes:

$\lim_{{x \to 1}} \frac{\frac{1 - x}{2(x+1)}}{x - 1}$

Step 2: Simplify the overall expression

We can rewrite this as:

$\lim_{{x \to 1}} \frac{1 - x}{2(x + 1)(x - 1)}$

Step 3: Factor the numerator

Notice that $1 - x = -(x - 1)$, so the expression becomes:

$\lim_{{x \to 1}} \frac{-(x - 1)}{2(x + 1)(x - 1)}$

We can now cancel out the $x - 1$ term in both the numerator and denominator, leaving us with:

$\lim_{{x \to 1}} \frac{-1}{2(x + 1)}$

Step 4: Evaluate the limit as $x \to 1$

Now we substitute $x = 1$ into the simplified expression:

$\frac{-1}{2(1 + 1)} = \frac{-1}{4}$

Thus, the value of the limit is:

$\boxed{-\frac{1}{4}}$

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Would you like further details on any step or more explanation? Here are 5 related questions to expand your understanding:

1. How does the factorization $1 - x = -(x - 1)$ help simplify limits?
2. Can this limit be evaluated using L'Hôpital's Rule?
3. What is the significance of finding common denominators when simplifying rational expressions?
4. How does canceling terms affect the evaluation of limits?
5. How do you handle limits involving division by zero?

Tip: Always check for opportunities to factor expressions when simplifying limits; this often helps eliminate terms that would otherwise make the limit undefined.