1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{x^2 - \frac{1}{9}}{x - 3}$$. 2. **Recall the formula and rules:** When direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we try to simplify the expression. 3. **Substitute $x=3$ directly:** $$\frac{3^2 - \frac{1}{9}}{3 - 3} = \frac{9 - \frac{1}{9}}{0} = \frac{\frac{81}{9} - \frac{1}{9}}{0} = \frac{\frac{80}{9}}{0}$$ which is undefined, so simplification is needed. 4. **Rewrite numerator:** $$x^2 - \frac{1}{9} = \left(x - \frac{1}{3}\right)\left(x + \frac{1}{3}\right)$$ (difference of squares). 5. **Rewrite the limit:** $$\lim_{x \to 3} \frac{\left(x - \frac{1}{3}\right)\left(x + \frac{1}{3}\right)}{x - 3}$$. 6. **Notice that direct substitution still gives division by zero, so check if numerator can be factored to cancel denominator:** Since $x - 3$ is not a factor of numerator, try to rewrite numerator as a single fraction: $$x^2 - \frac{1}{9} = \frac{9x^2 - 1}{9}$$. 7. **Rewrite the limit as:** $$\lim_{x \to 3} \frac{\frac{9x^2 - 1}{9}}{x - 3} = \lim_{x \to 3} \frac{9x^2 - 1}{9(x - 3)}$$. 8. **Factor numerator $9x^2 - 1$ as difference of squares:** $$9x^2 - 1 = (3x - 1)(3x + 1)$$. 9. **So the limit becomes:** $$\lim_{x \to 3} \frac{(3x - 1)(3x + 1)}{9(x - 3)}$$. 10. **Try to factor denominator to cancel with numerator:** Rewrite $x - 3$ as is, no common factor with numerator. 11. **Since direct substitution still gives division by zero, use algebraic manipulation:** Rewrite numerator and denominator to express in terms of $(x - 3)$: $$3x - 1 = 3(x - 3) + 8$$ $$3x + 1 = 3(x - 3) + 10$$ 12. **Substitute back:** $$\frac{(3(x - 3) + 8)(3(x - 3) + 10)}{9(x - 3)}$$. 13. **Expand numerator:** $$= \frac{9(x - 3)^2 + 54(x - 3) + 80}{9(x - 3)}$$. 14. **Split the fraction:** $$= \frac{9(x - 3)^2}{9(x - 3)} + \frac{54(x - 3)}{9(x - 3)} + \frac{80}{9(x - 3)} = (x - 3) + 6 + \frac{80}{9(x - 3)}$$. 15. **As $x \to 3$, $(x - 3) \to 0$, but $\frac{80}{9(x - 3)}$ diverges, so limit does not exist as finite number.** 16. **Re-examine original problem:** The original expression is: $$\frac{x^2 - \frac{1}{9}}{x - 3}$$ At $x=3$, numerator is $9 - \frac{1}{9} = \frac{80}{9}$, denominator is 0, so limit tends to infinity or negative infinity depending on direction. 17. **Therefore, the limit does not exist (infinite discontinuity).** **Final answer:** The limit does not exist because the denominator approaches zero while numerator approaches a nonzero constant.