1. Problem: Find $$\lim_{x \to 3} (2x + 1)$$ Solution: Substitute $x=3$ directly since the function is polynomial and continuous. $$2(3) + 1 = 6 + 1 = 7$$ 2. Problem: Compute $$\lim_{x \to 0} \frac{\sin x}{x}$$ This is a standard trigonometric limit. $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ using Squeeze Theorem. 3. Problem: Evaluate $$\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$$ Factor numerator: $$x^2 - 1 = (x-1)(x+1)$$. Cancel $(x-1)$: $$\lim_{x \to 1} (x+1) = 2$$ 4. Problem: $$\lim_{x \to \infty} \frac{5x^2 + 3}{2x^2 - x}$$ Divide numerator and denominator by $x^2$: $$\lim_{x \to \infty} \frac{5 + \frac{3}{x^2}}{2 - \frac{1}{x}} = \frac{5}{2}$$ 5. Problem: $$\lim_{x \to 0} \frac{e^x - 1}{x}$$ Recognize derivative of $e^x$ at 0; $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$ 6. Problem: Calculate $$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$$ Multiply numerator and denominator by conjugate $\sqrt{x} + 2$: $$\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x-4)(\sqrt{x} + 2)} = \frac{x - 4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x} + 2}$$ Substitute $x=4$: $$\frac{1}{2+2} = \frac{1}{4}$$ 7. Problem: $$\lim_{x \to 0^+} \ln(x)$$ As $x$ approaches 0 from the right, logarithm tends to $-\infty$. So $$\lim_{x \to 0^+} \ln(x) = -\infty$$ 8. Problem: $$\lim_{x \to 0} \frac{1 - \cos x}{x^2}$$ Use Taylor expansion: $1 - \cos x \approx \frac{x^2}{2}$. So limit is $$\lim_{x \to 0} \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2}$$ 9. Problem: $$\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$$ Factor numerator: $x^3 - 8 = (x-2)(x^2+2x+4)$. Cancel $(x-2)$: $$\lim_{x \to 2} (x^2 + 2x + 4) = 4 + 4 + 4 =12$$ 10. Problem: $$\lim_{x \to 0} \frac{\tan x}{x}$$ Using small angle approximation, limit is $$1$$ 11. Problem: $$\lim_{x \to 0} \frac{\sin 3x}{x}$$ Rewrite as $$3 \frac{\sin 3x}{3x}$$; $$3 \times 1 = 3$$ 12. Problem: $$\lim_{x \to \infty} (1 + \frac{1}{x})^x$$ Known definition of $e$; Limit is $$e$$ 13. Problem: $$\lim_{x \to 0} \frac{e^{2x} - 1}{x}$$ Use derivative of $e^{2x}$ at 0: $$2$$ 14. Problem: Find $$\lim_{x \to 0} \frac{\ln(1+x)}{x}$$ Using derivative of $\ln(1+x)$ at 0: $$1$$ 15. Problem: $$\lim_{x \to 1} \frac{\ln x}{x-1}$$ Derivative of $\ln x$ at 1: $$1$$ 16. Problem: $$\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$$ Multiply numerator and denominator by conjugate: $$\frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)} = \frac{x}{x(\sqrt{1+x} + 1)} = \frac{1}{\sqrt{1+x} + 1}$$ Substitute $x=0$: $$\frac{1}{2}$$ 17. Real-life application: Velocity Problem: The instantaneous velocity of a particle with position function $s(t)=t^2$ at time $t=3$ is $$\lim_{h \to 0} \frac{(3+h)^2 - 3^2}{h}$$ Expand numerator: $$\frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h$$ Taking limit as $h\to0$: velocity is $$6$$ 18. Problem: $$\lim_{x \to 0^+} x \ln x$$ Rewrite as $$\frac{\ln x}{1/x}$$, both go to infinity with opposite signs Using L'Hôpital's rule: $$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$$ 19. Problem: $$\lim_{x \to 2^-} \frac{1}{x-2}$$ As $x$ approaches 2 from left, denominator approaches zero negative side, Limit is $$-\infty$$ 20. Problem: $$\lim_{x \to 0} \frac{1 - e^{-x}}{x}$$ Using derivative of $e^{-x}$ at 0: $$1$$ Summary: Multiple limits calculated using substitution, factorization, conjugates, L'Hôpital's Rule, Squeeze Theorem, and known limits.