1. Problem: Find the limit $$\lim_{x \to -7} (2x + 5)$$ Step 1: Substitute $x = -7$ directly since the function is linear and continuous. Step 2: Calculate $2(-7) + 5 = -14 + 5 = -9$ Answer: $$\lim_{x \to -7} (2x + 5) = -9$$ 2. Problem: Find the limit $$\lim_{x \to 7^-} (2x + 5)$$ Step 1: Since the function is continuous, the left-hand limit equals the function value at 7. Step 2: Calculate $2(7) + 5 = 14 + 5 = 19$ Answer: $$\lim_{x \to 7^-} (2x + 5) = 19$$ 3. Problem: Find the limit $$\lim_{x \to -\infty} (2x + 5)$$ Step 1: As $x$ approaches negative infinity, $2x$ dominates. Step 2: Since $2x \to -\infty$, the whole expression tends to $-\infty$. Answer: $$\lim_{x \to -\infty} (2x + 5) = -\infty$$ 4. Problem: Find the limit $$\lim_{x \to -4} (x + 3)^{2006}$$ Step 1: Substitute $x = -4$ directly. Step 2: Calculate $(-4 + 3)^{2006} = (-1)^{2006}$. Step 3: Since 2006 is even, $(-1)^{2006} = 1$. Answer: $$\lim_{x \to -4} (x + 3)^{2006} = 1$$ 5. Problem: Find the limit $$\lim_{x \to -4} (x + 3)^{2007}$$ Step 1: Substitute $x = -4$. Step 2: Calculate $(-1)^{2007}$. Step 3: Since 2007 is odd, $(-1)^{2007} = -1$. Answer: $$\lim_{x \to -4} (x + 3)^{2007} = -1$$ 6. Problem: Find the limit $$\lim_{x \to -\infty} (x + 3)^{2007}$$ Step 1: For large negative $x$, $x + 3 \approx x$ which is negative. Step 2: Since 2007 is odd, $(x + 3)^{2007}$ preserves the sign of $x + 3$. Step 3: As $x \to -\infty$, $(x + 3)^{2007} \to -\infty$. Answer: $$\lim_{x \to -\infty} (x + 3)^{2007} = -\infty$$ 7. Problem: Find the limit $$\lim_{t \to 1} \frac{t^2 + t - 2}{t^2 - 1}$$ Step 1: Factor numerator and denominator: $$t^2 + t - 2 = (t + 2)(t - 1)$$ $$t^2 - 1 = (t - 1)(t + 1)$$ Step 2: Simplify the expression: $$\frac{(t + 2)(t - 1)}{(t - 1)(t + 1)} = \frac{t + 2}{t + 1}, \quad t \neq 1$$ Step 3: Substitute $t = 1$: $$\frac{1 + 2}{1 + 1} = \frac{3}{2}$$ Answer: $$\lim_{t \to 1} \frac{t^2 + t - 2}{t^2 - 1} = \frac{3}{2}$$ 8. Problem: Find the limit $$\lim_{t \to 1^+} \frac{t^2 + t - 2}{t^2 - 1}$$ Step 1: Using the simplified form from previous step: $$\frac{t + 2}{t + 1}$$ Step 2: Substitute $t = 1$: $$\frac{3}{2}$$ Answer: $$\lim_{t \to 1^+} \frac{t^2 + t - 2}{t^2 - 1} = \frac{3}{2}$$ 9. Problem: Find the limit $$\lim_{t \to -1} \frac{t^2 + t - 2}{t^2 - 1}$$ Step 1: Factor as before: $$\frac{(t + 2)(t - 1)}{(t - 1)(t + 1)} = \frac{t + 2}{t + 1}$$ for $t \neq \pm 1$ Step 2: Substitute $t = -1$: Denominator $t + 1 = 0$, so direct substitution is undefined. Step 3: Check left and right limits: - As $t \to -1^-$, denominator approaches 0 from negative side, numerator $t + 2 = 1$. - As $t \to -1^+$, denominator approaches 0 from positive side. Step 4: Limits approach $\pm \infty$; limit does not exist. Answer: $$\lim_{t \to -1} \frac{t^2 + t - 2}{t^2 - 1} \text{ does not exist}$$ 10. Problem: Find the limit $$\lim_{x \to \infty} \frac{x^2 + 3}{x^2 + 4}$$ Step 1: Divide numerator and denominator by $x^2$: $$\frac{1 + \frac{3}{x^2}}{1 + \frac{4}{x^2}}$$ Step 2: As $x \to \infty$, terms with $\frac{1}{x^2} \to 0$. Step 3: Limit is $$\frac{1 + 0}{1 + 0} = 1$$ Answer: $$\lim_{x \to \infty} \frac{x^2 + 3}{x^2 + 4} = 1$$ 11. Problem: Find the limit $$\lim_{x \to \infty} \frac{x^5 + 3}{x^2 + 4}$$ Step 1: Highest power in numerator is $x^5$, denominator $x^2$. Step 2: For large $x$, numerator grows faster than denominator. Step 3: Limit tends to $\infty$. Answer: $$\lim_{x \to \infty} \frac{x^5 + 3}{x^2 + 4} = \infty$$ 12. Problem: Find the limit $$\lim_{x \to \infty} \frac{x^2 + 1}{x^5 + 2}$$ Step 1: Highest power in denominator is $x^5$, numerator $x^2$. Step 2: For large $x$, denominator grows faster. Step 3: Limit tends to 0. Answer: $$\lim_{x \to \infty} \frac{x^2 + 1}{x^5 + 2} = 0$$ 13. Problem: Find the limit $$\lim_{x \to \infty} \frac{(2x + 1)^4}{(3x^2 + 1)^2}$$ Step 1: Expand dominant terms: Numerator: $(2x)^4 = 16x^4$ Denominator: $(3x^2)^2 = 9x^4$ Step 2: For large $x$, expression behaves like: $$\frac{16x^4}{9x^4} = \frac{16}{9}$$ Answer: $$\lim_{x \to \infty} \frac{(2x + 1)^4}{(3x^2 + 1)^2} = \frac{16}{9}$$ 14. Problem: Find the limit $$\lim_{u \to \infty} \frac{(2u + 1)^4}{(3u^2 + 1)^2}$$ Step 1: Same as previous problem, replace $x$ by $u$. Step 2: Limit is $$\frac{16}{9}$$ Answer: $$\lim_{u \to \infty} \frac{(2u + 1)^4}{(3u^2 + 1)^2} = \frac{16}{9}$$ 15. Problem: Find the limit $$\lim_{t \to 0} \frac{(2t + 1)^4}{(3t^2 + 1)^2}$$ Step 1: Substitute $t = 0$ directly: Numerator: $(2(0) + 1)^4 = 1^4 = 1$ Denominator: $(3(0)^2 + 1)^2 = 1^2 = 1$ Step 2: Limit is $$\frac{1}{1} = 1$$ Answer: $$\lim_{t \to 0} \frac{(2t + 1)^4}{(3t^2 + 1)^2} = 1$$ 16. Problem: Coordinates of points A,...,E on graph $y = \sin \frac{\pi}{x}$ Step 1: Points labeled likely correspond to specific $x$ values. Step 2: For example, at $x = 1$, $y = \sin \pi = 0$. Step 3: At $x = \frac{1}{2}$, $y = \sin 2\pi = 0$. Step 4: At $x = \frac{1}{4}$, $y = \sin 4\pi = 0$. Step 5: At $x = \frac{2}{3}$, $y = \sin \frac{3\pi}{2} = -1$. Step 6: At $x = \frac{1}{3}$, $y = \sin 3\pi = 0$. Answer: Coordinates depend on exact points but follow $y = \sin \frac{\pi}{x}$. 17. Problem: If $$\lim_{x \to a} f(x)$$ exists then $f$ is continuous at $x = a$. True or false? Step 1: Continuity at $x = a$ requires: $$\lim_{x \to a} f(x) = f(a)$$ Step 2: Limit existing does not guarantee $f(a)$ equals the limit. Answer: False 18. Problem: Give two examples where $$\lim_{x \to 0} f(x)$$ does not exist. Example 1: $f(x) = \frac{1}{x}$ Example 2: $f(x) = \sin \frac{1}{x}$ 19. Problem: If $$\lim_{x \to 0} f(x)$$ and $$\lim_{x \to 0} g(x)$$ both do not exist, then $$\lim_{x \to 0} (f(x) + g(x))$$ also does not exist. True or false? Answer: False (sum may exist even if individual limits do not) 20. Problem: If $$\lim_{x \to 0} f(x)$$ and $$\lim_{x \to 0} g(x)$$ both do not exist, then $$\lim_{x \to 0} \frac{f(x)}{g(x)}$$ also does not exist. True or false? Answer: False (quotient limit may exist in some cases) 21. Problem: Show that $$\lim_{x \to \infty} x$$ does not exist given $$\lim_{x \to \infty} \frac{1}{x} = 0$$. Step 1: Suppose $$\lim_{x \to \infty} x = L$$ for some finite $L$. Step 2: Then $$\lim_{x \to \infty} x \cdot \frac{1}{x} = \lim_{x \to \infty} 1 = 1$$. Step 3: By limit properties, this equals $$L \cdot 0 = 0$$. Step 4: Contradiction since 1 \neq 0. Answer: $$\lim_{x \to \infty} x$$ does not exist. 22. Problem: Evaluate $$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$$ Step 1: Multiply numerator and denominator by $$\sqrt{x} + 3$$: $$\frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} = \frac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \frac{1}{\sqrt{x} + 3}$$ Step 2: Substitute $x = 9$: $$\frac{1}{3 + 3} = \frac{1}{6}$$ Answer: $$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} = \frac{1}{6}$$