1. **Problem 1:** Find the limit $$\lim_{x \to 0} \left(1 + \frac{1}{x}\right)^x$$. 2. **Step 1:** Recognize that as $$x \to 0$$, the expression inside the parentheses $$1 + \frac{1}{x}$$ becomes unbounded, and the exponent $$x$$ approaches 0. This is an indeterminate form of type $$1^\infty$$. 3. **Step 2:** Use the substitution $$t = \frac{1}{x}$$, so as $$x \to 0$$, $$t \to \pm \infty$$ depending on the direction. 4. **Step 3:** Rewrite the limit as $$\lim_{t \to \pm \infty} \left(1 + t\right)^{\frac{1}{t}}$$. 5. **Step 4:** For $$t \to +\infty$$, $$1 + t \to +\infty$$ and the exponent $$\frac{1}{t} \to 0^+$$. The expression behaves like $$\infty^0$$, which is indeterminate. 6. **Step 5:** Consider the natural logarithm: $$\ln y = \frac{1}{t} \ln(1 + t)$$. 7. **Step 6:** As $$t \to +\infty$$, $$\ln(1 + t) \sim \ln t$$, so $$\ln y \sim \frac{\ln t}{t}$$. 8. **Step 7:** Since $$\lim_{t \to +\infty} \frac{\ln t}{t} = 0$$, we have $$\ln y \to 0$$, so $$y \to e^0 = 1$$. 9. **Step 8:** For $$t \to -\infty$$, $$1 + t$$ is negative and the expression is not defined for real numbers, so the limit from the left does not exist. 10. **Answer 1:** The limit $$\lim_{x \to 0^+} \left(1 + \frac{1}{x}\right)^x = 1$$, and the limit from the left does not exist. --- 11. **Problem 2:** Find the limit $$\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$$. 12. **Step 1:** Notice direct substitution gives $$\frac{1 - 1}{1 - 1} = \frac{0}{0}$$, an indeterminate form. 13. **Step 2:** Factor numerator: $$x^2 - 1 = (x - 1)(x + 1)$$. 14. **Step 3:** Simplify the expression: $$\frac{(x - 1)(x + 1)}{x - 1} = x + 1, \quad x \neq 1$$. 15. **Step 4:** Now substitute $$x = 1$$: $$1 + 1 = 2$$. 16. **Answer 2:** The limit is $$2$$. --- 17. **Problem 3:** Find the limit $$\lim_{x \to 3} \frac{|x - 3|}{x - 3}$$. 18. **Step 1:** Consider the left-hand limit $$x \to 3^-$$: When $$x < 3$$, $$x - 3 < 0$$, so $$|x - 3| = -(x - 3)$$. 19. **Step 2:** Substitute: $$\frac{|x - 3|}{x - 3} = \frac{-(x - 3)}{x - 3} = -1$$. 20. **Step 3:** Consider the right-hand limit $$x \to 3^+$$: When $$x > 3$$, $$x - 3 > 0$$, so $$|x - 3| = x - 3$$. 21. **Step 4:** Substitute: $$\frac{|x - 3|}{x - 3} = \frac{x - 3}{x - 3} = 1$$. 22. **Step 5:** Since the left-hand limit is $$-1$$ and the right-hand limit is $$1$$, the two-sided limit does not exist. 23. **Answer 3:** The limit does not exist because the left and right limits differ.