Use the intuitive approach to explain why the limit $$ \lim_{x\rarr2\text{ }} $$$$ \frac{x^2-4}{x-2} $$exists, and compute the limit.

Let's explore the limit using an intuitive approach.

We are asked to find the limit:

\[
\lim_{x \to 2} \frac{x^2 - 4}{x - 2}
\]

### Step 1: Substitute $ x = 2 $
First, let's try plugging $ x = 2 $ directly into the expression:

\[
\frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{0} = \frac{0}{0}
\]

This results in the indeterminate form $ \frac{0}{0} $. It means we cannot directly compute the limit by simple substitution, but it also suggests that we might be able to simplify the expression.

### Step 2: Simplify the expression
The numerator, $ x^2 - 4 $, is a difference of squares. We can factor it as:

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

So, the original expression becomes:

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

Now, for $ x \neq 2 $, we can cancel the common factor $ x - 2 $ from both the numerator and the denominator:

\[
\frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for} \quad x \neq 2
\]

### Step 3: Evaluate the limit of the simplified expression
Now that we have simplified the expression to $ x + 2 $, we can substitute $ x = 2 $ without any issues:

\[
\lim_{x \to 2} (x + 2) = 2 + 2 = 4
\]

### Conclusion:
Therefore, the limit exists and its value is:

\[
\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4
\]

### Why does the limit exist?
The indeterminate form $ \frac{0}{0} $ indicates that the function has a removable discontinuity at $ x = 2 $. After simplifying the expression, the discontinuity is "removed," and we are left with a continuous function $ x + 2 $, which has no problems at $ x = 2 $. This is why the limit exists and equals 4.

---

Would you like further details or have any questions?

Here are 5 related questions:
1. How do we handle other indeterminate forms like $ \frac{0}{0} $?
2. What is the difference between a removable and non-removable discontinuity?
3. How does factoring help in solving limits involving polynomials?
4. Can limits be used to define continuity of a function?
5. How does L'Hopital's rule apply to indeterminate forms like $ \frac{0}{0} $?

**Tip:** When dealing with indeterminate forms, always look for algebraic simplifications such as factoring, rationalizing, or canceling common terms.

Use the intuitive approach to explain why the limit $$ \lim_{x \to 2} \frac{x^2-4}{x-2} $$ exists, and compute the limit.

Learn how to compute the limit of (x^2-4)/(x-2) as x approaches 2 using an intuitive approach. This problem covers key concepts in limits, factoring, and handling indeterminate forms, suitable for students in Grades 11-12 or college calculus.

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