The given limit expression is:

$\lim_{{x \to \frac{1}{2}}} \left( \frac{1}{x - 1} \right) \left( \frac{1}{x + 3} - \frac{2x}{3x + 5} \right)$

Step-by-Step Solution

1. Analyze the Limit

We are tasked with evaluating the limit as $x \to \frac{1}{2}$. First, we should try directly substituting $x = \frac{1}{2}$ into the expression to see if there are any indeterminate forms or if it simplifies.

2. Direct Substitution of $x = \frac{1}{2}$

• For the first term $\frac{1}{x - 1}$, substituting $x = \frac{1}{2}$: $\frac{1}{\frac{1}{2} - 1} = \frac{1}{-\frac{1}{2}} = -2$

• Now for the second term $\frac{1}{x + 3} - \frac{2x}{3x + 5}$, let's break it down:

  • For $\frac{1}{x + 3}$ when $x = \frac{1}{2}$: $\frac{1}{\frac{1}{2} + 3} = \frac{1}{\frac{7}{2}} = \frac{2}{7}$
  • For $\frac{2x}{3x + 5}$ when $x = \frac{1}{2}$: $\frac{2 \times \frac{1}{2}}{3 \times \frac{1}{2} + 5} = \frac{1}{\frac{3}{2} + 5} = \frac{1}{\frac{13}{2}} = \frac{2}{13}$ So, the second term becomes: $\frac{2}{7} - \frac{2}{13}$ To subtract these, find the common denominator: $\frac{2}{7} - \frac{2}{13} = \frac{26}{91} - \frac{14}{91} = \frac{12}{91}$

3. Combine the Two Terms

Now, combining the two terms: $-2 \times \frac{12}{91} = \frac{-24}{91}$

Thus, the limit is:

$\lim_{{x \to \frac{1}{2}}} \left( \frac{1}{x - 1} \right) \left( \frac{1}{x + 3} - \frac{2x}{3x + 5} \right) = \frac{-24}{91}$

Final Answer:

$\boxed{\frac{-24}{91}}$

Would you like further details or clarifications on any step?

Here are five related questions to deepen your understanding:

1. How do we identify indeterminate forms in limits?
2. What strategies can be used if direct substitution results in an indeterminate form?
3. How do we subtract rational expressions with different denominators?
4. Can L'Hopital's Rule be applied to this limit if an indeterminate form occurs?
5. How does multiplying two terms affect the overall result in a limit?

Tip: Always check for indeterminate forms when calculating limits directly to decide whether simplification or other techniques are needed.