To prove that the function $f(x) = x^2$ is continuous at $x = 3$ using the definition of continuity, we need to show that for any given $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x - 3| < \delta$, it follows that $|f(x) - f(3)| < \epsilon$.

Step-by-Step Proof:

1. Definition of Continuity: A function $f(x)$ is continuous at $x = 3$ if: $\lim_{x \to 3} f(x) = f(3)$ This is equivalent to saying that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x - 3| < \delta$, then $|f(x) - f(3)| < \epsilon$.

2. Substitute the Function $f(x) = x^2$: We need to find $\delta$ such that: $|f(x) - f(3)| = |x^2 - 9| < \epsilon$

3. Simplify the Expression: We can factorize $|x^2 - 9|$ as: $|x^2 - 9| = |(x - 3)(x + 3)|$ So, the inequality we need to satisfy becomes: $|(x - 3)(x + 3)| < \epsilon$

4. Establish a Bound on $|x + 3|$: To find a $\delta$, note that if $|x - 3| < \delta$, then $x$ is close to 3. Specifically, we can bound $|x + 3|$: $|x + 3| \leq |3 + 3| + |x - 3| = 6 + \delta$ This means $|(x - 3)(x + 3)| \leq \delta(6 + \delta)$.

5. Choose $\delta$: We want $\delta$ to satisfy: $\delta(6 + \delta) < \epsilon$ To simplify the choice of $\delta$, we can choose $\delta$ small enough so that $6 + \delta \approx 6$. This approximation gives: $\delta \times 6 < \epsilon \quad \Rightarrow \quad \delta < \frac{\epsilon}{6}$

6. Conclusion: Thus, for any given $\epsilon > 0$, we can choose $\delta = \min\left(1, \frac{\epsilon}{6}\right)$. This ensures that whenever $|x - 3| < \delta$, the condition $|x^2 - 9| < \epsilon$ is satisfied.

Therefore, by the definition of continuity, $f(x) = x^2$ is continuous at $x = 3$.

Do you have any questions or need further details?

5 Related Questions:

1. How would you prove continuity for other points, such as $x = 0$?
2. What is the geometric interpretation of continuity for polynomial functions?
3. Can you apply this method to non-polynomial functions?
4. How does this proof differ when proving differentiability at a point?
5. What role does the choice of $\delta$ play in determining the continuity of a function?

Tip:

When proving continuity, carefully choose $\delta$ in relation to $\epsilon$ and make sure it holds for the entire domain around the point of interest.