$$ \lim_{x\to3} $$$$ \frac{\left(3+x^{}\right)^2-3^2}{\left(x-3\right)} $$

The given limit is:

\[
\lim_{x \to 3} \frac{(3 + x)^2 - 3^2}{x - 3}
\]

First, simplify the numerator:

\[
(3 + x)^2 - 3^2 = [(3 + x) - 3][(3 + x) + 3]
\]
This follows from the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$.

Now, apply this:

\[
(3 + x)^2 - 3^2 = (x)(x + 6)
\]

Thus, the expression becomes:

\[
\lim_{x \to 3} \frac{x(x + 6)}{x - 3}
\]

Since direct substitution leads to $0/0$, we can factor and cancel terms. Rewrite the expression:

\[
\lim_{x \to 3} \frac{x(x + 6)}{x - 3}
\]

This cannot be simplified directly. Instead, let's re-examine the structure or use L'Hopital's Rule, if necessary.

Do you need further details or have questions?

---
**Questions for exploration:**
1. How is L'Hopital's Rule used in limit problems?
2. What is the difference of squares formula?
3. How do you factor higher-degree polynomials?
4. How can we handle indeterminate forms in limits?
5. How does substitution work in limits?

**Tip:** When encountering $0/0$ forms, always consider factoring or applying L'Hopital’s Rule.

Evaluate $$ \lim_{x \to 3} \frac{(3 + x)^2 - 3^2}{x - 3} $$

Learn how to evaluate the limit $ \lim_{x \to 3} \frac{(3 + x)^2 - 3^2}{x - 3} $ using the difference of squares method and L'Hopital's Rule. This solution explores factoring techniques, indeterminate forms, and the application of L'Hopital's Rule. Ideal for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation