\lim _{x\to \:2}\left(\frac{x^2-4}{x-2}\right)

The limit you're trying to evaluate is:

\[
\lim_{x \to 2} \left(\frac{x^2 - 4}{x - 2}\right)
\]

First, notice that the expression \( x^2 - 4 \) can be factored as a difference of squares:

\[
x^2 - 4 = (x - 2)(x + 2)
\]

So, the expression becomes:

\[
\lim_{x \to 2} \left(\frac{(x - 2)(x + 2)}{x - 2}\right)
\]

Since \( x - 2 \) appears in both the numerator and the denominator, we can cancel it out (for \( x \neq 2 \)):

\[
\lim_{x \to 2} (x + 2)
\]

Now, simply substitute \( x = 2 \) into the expression:

\[
2 + 2 = 4
\]

So, the limit is:

\[
\lim_{x \to 2} \left(\frac{x^2 - 4}{x - 2}\right) = 4
\]

Would you like further details or have any other questions?

Here are 5 related questions for practice:
1. What is the limit \(\lim_{x \to 3} \left(\frac{x^2 - 9}{x - 3}\right)\)?
2. How do you find the limit of a function when direct substitution results in an indeterminate form?
3. Can all limits be evaluated using algebraic simplification, or are there exceptions?
4. What is the limit \(\lim_{x \to -1} \left(\frac{x^2 - 1}{x + 1}\right)\)?
5. How does L'Hôpital's rule apply to limits that result in the \(\frac{0}{0}\) indeterminate form?

**Tip:** Always check if a function can be simplified before applying more advanced limit techniques like L'Hôpital's rule.

Understand how to evaluate the limit of (x^2 - 4) / (x - 2) as x approaches 2 with a step-by-step solution. Learn about factoring, algebraic simplification, and handling indeterminate forms in limits. Suitable for advanced high school students and anyone interested in calculus and algebra.

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