The limit presented is:

$\lim_{x \to 2} \left( \frac{1}{x-2} - \frac{4}{x^2 - 4} \right)$

Step-by-Step Solution:

Step 1: Factor the denominator in the second term

The denominator $x^2 - 4$ can be factored as:

$x^2 - 4 = (x - 2)(x + 2)$

So the second term becomes:

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

Thus, the limit now looks like:

$\lim_{x \to 2} \left( \frac{1}{x-2} - \frac{4}{(x-2)(x+2)} \right)$

Step 2: Combine the two terms into a single fraction

To combine the two fractions, the common denominator will be $(x-2)(x+2)$. The first fraction becomes:

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

Now, we subtract the second term:

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

Simplifying the numerator:

$(x + 2) - 4 = x - 2$

So the expression becomes:

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

Step 3: Cancel out the common factor

The $x - 2$ cancels out from the numerator and denominator, leaving:

$\lim_{x \to 2} \frac{1}{x + 2}$

Step 4: Evaluate the limit

Now, substitute $x = 2$:

$\frac{1}{2 + 2} = \frac{1}{4}$

Thus, the value of the limit is:

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

Would you like further clarification on any step or additional details?

Related Questions:

1. How can you evaluate a limit involving a rational expression with removable discontinuities?
2. What is the difference between a removable and non-removable discontinuity?
3. How can factoring help simplify expressions in limits involving polynomials?
4. What are the common techniques to solve limits of indeterminate forms?
5. How do you handle limits approaching infinity?

Tip: Always check for possible factorizations when faced with polynomials in limits—this can often simplify the problem greatly!