1. **Problem:** Evaluate $$\lim\limits_{(x,y)\to(0,\ln 2)} e^{x-y}$$ Step 1: Substitute $x=0$ and $y=\ln 2$ directly since exponential is continuous. Step 2: Compute exponent: $$0 - \ln 2 = -\ln 2$$ Step 3: Evaluate exponential: $$e^{-\ln 2} = \frac{1}{e^{\ln 2}} = \frac{1}{2}$$ Answer: $$\boxed{\frac{1}{2}}$$ 2. **Problem:** Evaluate $$\lim\limits_{(x,y)\to(1,1)} \ln|1 + x^2 y^2|$$ Step 1: Substitute $x=1$, $y=1$. Step 2: Compute inside logarithm: $$1 + (1)^2(1)^2 = 1 + 1 = 2$$ Step 3: Take natural log: $$\ln 2$$ Answer: $$\boxed{\ln 2}$$ 3. **Problem:** Evaluate $$\lim\limits_{(x,y)\to(0,0)} \frac{e^y \sin x}{x}$$ Step 1: Consider limit as $(x,y)\to(0,0)$. Step 2: Since $\sin x / x \to 1$ as $x \to 0$, and $e^y \to 1$ as $y \to 0$, Step 3: Combine limits: $$\lim e^y \cdot \lim \frac{\sin x}{x} = 1 \cdot 1 = 1$$ Answer: $$\boxed{1}$$ 4. **Problem:** Evaluate $$\lim\limits_{(x,y)\to(\frac{1}{2},7)} \cos \sqrt[3]{xy}$$ Step 1: Substitute $x=1/2$, $y=7$. Step 2: Compute cube root term: $$\sqrt[3]{\frac{1}{2} \times 7} = \sqrt[3]{\frac{7}{2}}$$ Step 3: Evaluate cosine: $$\cos \left( \sqrt[3]{\frac{7}{2}} \right)$$ Answer: $$\boxed{\cos \left( \sqrt[3]{\frac{7}{2}} \right)}$$ 5. **Problem:** Evaluate $$\lim\limits_{(x,y)\to(1, \frac{\pi}{6})} \frac{x \sin y}{x^2 +1}$$ Step 1: Substitute $x=1$, $y=\frac{\pi}{6}$. Step 2: Compute numerator: $$1 \times \sin \frac{\pi}{6} = \frac{1}{2}$$ Step 3: Compute denominator: $$1^2 + 1 = 2$$ Step 4: Evaluate fraction: $$\frac{\frac{1}{2}}{2} = \frac{1}{4}$$ Answer: $$\boxed{\frac{1}{4}}$$ 6. **Problem:** Evaluate $$\lim\limits_{(x,y)\to(\frac{\pi}{2}, 0)} \frac{\cos y +1}{y - \sin x}$$ Step 1: Substitute $x=\frac{\pi}{2}$ and $y=0$. Step 2: Numerator: $$\cos 0 + 1 = 1 + 1 = 2$$ Step 3: Denominator: $$0 - \sin \frac{\pi}{2} = 0 - 1 = -1$$ Step 4: Compute limit: $$\frac{2}{-1} = -2$$ Answer: $$\boxed{-2}$$ 13. **Problem:** $$\lim\limits_{(x,y)\to(1,1), x \neq y} \frac{x^2 - 2xy + y^2}{x - y}$$ Step 1: Recognize numerator as perfect square: $$x^2 - 2xy + y^2 = (x - y)^2$$ Step 2: Simplify: $$\frac{(x - y)^2}{x - y} = x - y$$ Step 3: Take limit: $$(1) - (1) = 0$$ Answer: $$\boxed{0}$$ 14. **Problem:** $$\lim\limits_{(x,y)\to(1,1), x \neq y} \frac{x^2 - y^2}{x - y}$$ Step 1: Factor numerator: $$x^2 - y^2 = (x - y)(x + y)$$ Step 2: Simplify: $$\frac{(x - y)(x + y)}{x - y} = x + y$$ Step 3: Evaluate limit: $$1 + 1 = 2$$ Answer: $$\boxed{2}$$ 15. **Problem:** $$\lim\limits_{(x,y)\to(1,1), x\neq1} \frac{xy - y - 2x + 2}{x - 1}$$ Step 1: Factor numerator grouping: $$xy - y - 2x + 2 = y(x - 1) - 2(x - 1) = (x - 1)(y - 2)$$ Step 2: Simplify: $$\frac{(x - 1)(y - 2)}{x - 1} = y - 2$$ Step 3: Evaluate limit: $$1 - 2 = -1$$ Answer: $$\boxed{-1}$$ 16. **Problem:** $$\lim\limits_{(x,y)\to(2,4), x \neq 4, x \neq x^2} \frac{y + 4}{x^2 y - xy + 4x^2 - 4x}$$ Step 1: Factor denominator terms by grouping: $$x^2 y - xy + 4x^2 - 4x = xy(x - 1) + 4x(x - 1) = (x - 1)(xy + 4x) = x(x - 1)(y + 4)$$ Step 2: Expression simplifies to: $$\frac{y + 4}{x(x - 1)(y + 4)} = \frac{1}{x(x - 1)}$$ Step 3: Substitute $x=2$: $$\frac{1}{2(2 - 1)} = \frac{1}{2}$$ Answer: $$\boxed{\frac{1}{2}}$$ 17. **Problem:** $$\lim\limits_{(x,y)\to(0,0), x \neq y} \frac{x - y + 2\sqrt{x} - 2\sqrt{y}}{\sqrt{x} - \sqrt{y}}$$ Step 1: Rewrite numerator: $$x - y + 2\sqrt{x} - 2\sqrt{y} = (x - y) + 2(\sqrt{x} - \sqrt{y})$$ Step 2: Express $x - y$ in terms of $\sqrt{x}, \sqrt{y}$: $$x - y = (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})$$ Step 3: Substitute: $$\frac{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) + 2(\sqrt{x} - \sqrt{y})}{\sqrt{x} - \sqrt{y}}$$ Step 4: Factor numerator: $$(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y} + 2)$$ Step 5: Simplify expression: $$\sqrt{x} + \sqrt{y} + 2$$ Step 6: Take limit as $(x,y)\to(0,0)$: $$0 + 0 + 2 = 2$$ Answer: $$\boxed{2}$$ 18. **Problem:** $$\lim\limits_{(x,y)\to(2,2), x+y\neq4} \frac{x + y -4}{\sqrt{x} + y -2}$$ Step 1: At $(2,2)$ numerator is 0; denominator is: $$\sqrt{2} + 2 - 2 = \sqrt{2}$$ Step 2: Since denominator $\neq 0$, the fraction goes to: $$\frac{0}{\sqrt{2}} = 0$$ Answer: $$\boxed{0}$$ 19. **Problem:** $$\lim\limits_{(x,y)\to(2,0), 2x - y \neq 4} \frac{\sqrt{2x} - y - 2}{2x - y - 4}$$ Step 1: Substitute $(2,0)$: Numerator: $$\sqrt{4} - 0 - 2 = 2 - 2 = 0$$ Denominator: $$4 - 0 - 4 = 0$$ Step 2: Indeterminate form; apply algebra. Step 3: Set $a=\sqrt{2x}$, rewrite numerator and denominator: Numerator: $$a - y - 2$$ Denominator: $$2x - y - 4 = (a)^2 - y - 4$$ Step 4: Since $a^2 - y -4 = (a - 2)(a + 2) - y$ But better approach: Consider approach along $y=0$, then Limit becomes: $$\lim_{x\to2} \frac{\sqrt{2x} - 2}{2x - 4} = \lim_{x \to 2} \frac{\sqrt{2x} - 2}{2(x - 2)}$$ Step 5: Use substitution $u = 2x$: $$\lim_{u \to 4} \frac{\sqrt{u} - 2}{2(\frac{u}{2} - 2)} = \lim_{u \to 4} \frac{\sqrt{u} - 2}{u - 4}$$ Step 6: Multiply numerator and denominator by conjugate: $$ \frac{\sqrt{u} - 2}{u - 4} \times \frac{\sqrt{u} + 2}{\sqrt{u} + 2} = \frac{u - 4}{(u - 4)(\sqrt{u} + 2)} = \frac{1}{\sqrt{u} + 2}$$ Step 7: Substitute $u=4$: $$ \frac{1}{2 + 2} = \frac{1}{4}$$ Answer: $$\boxed{\frac{1}{4}}$$ 20. **Problem:** $$\lim\limits_{(x,y)\to(4,3), x \neq y + 1} \frac{\sqrt{x} - \sqrt{y + 1}}{x - y - 1}$$ Step 1: At $(4,3)$ numerator: $$\sqrt{4} - \sqrt{3 + 1} = 2 - 2 = 0$$ Denominator: $$4 - 3 - 1 = 0$$ Indeterminate form. Step 2: Multiply numerator and denominator by conjugate: $$\frac{\sqrt{x} - \sqrt{y + 1}}{x - y - 1} \times \frac{\sqrt{x} + \sqrt{y + 1}}{\sqrt{x} + \sqrt{y + 1}} = \frac{x - (y+1)}{(x - y - 1)(\sqrt{x} + \sqrt{y + 1})} = \frac{x - y - 1}{(x - y - 1)(\sqrt{x} + \sqrt{y + 1})}$$ Step 3: Cancel $(x - y - 1)$: $$\frac{1}{\sqrt{x} + \sqrt{y + 1}}$$ Step 4: Substitute $(4,3)$: $$\frac{1}{2 + 2} = \frac{1}{4}$$ Answer: $$\boxed{\frac{1}{4}}$$ 21. **Problem:** $$\lim\limits_{(x,y)\to(0,0)} \frac{\sin (x^2 + y^2)}{x^2 + y^2}$$ Step 1: Let $r^2 = x^2 + y^2$, as $(x,y)\to(0,0)$, $r \to 0$. Step 2: Use single variable limit: $$\lim\limits_{r\to 0} \frac{\sin r^2}{r^2} = 1$$ Answer: $$\boxed{1}$$ 22. **Problem:** $$\lim\limits_{(x,y)\to(0,0)} \frac{1 - \cos(xy)}{xy}$$ Step 1: Let $t = xy$; as $(x,y)\to(0,0)$, $t \to 0$. Step 2: Use Taylor approximation: $$1 - \cos t \approx \frac{t^2}{2}$$ Step 3: Substitute: $$\frac{1 - \cos t}{t} \approx \frac{t^2 / 2}{t} = \frac{t}{2} \to 0$$ Answer: $$\boxed{0}$$ 23. **Problem:** $$\lim\limits_{(x,y)\to(1,-1)} \frac{x^3 + y^3}{x + y}$$ Step 1: Factor numerator using sum of cubes: $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$ Step 2: Simplify expression: $$\frac{(x + y)(x^2 - xy + y^2)}{x + y} = x^2 - xy + y^2$$ Step 3: Substitute $x=1$, $y=-1$: $$1^2 - 1 \times (-1) + (-1)^2 = 1 + 1 + 1 = 3$$ Answer: $$\boxed{3}$$ 24. **Problem:** $$\lim\limits_{(x,y)\to(2,2)} \frac{x - y}{x^4 - y^4}$$ Step 1: Factor denominator as difference of fourth powers: $$x^4 - y^4 = (x - y)(x + y)(x^2 + y^2)$$ Step 2: Simplify fraction: $$\frac{x - y}{(x - y)(x + y)(x^2 + y^2)} = \frac{1}{(x + y)(x^2 + y^2)}$$ Step 3: Substitute $(2,2)$: $$\frac{1}{(2 + 2)(2^2 + 2^2)} = \frac{1}{4 \times (4 + 4)} = \frac{1}{4 \times 8} = \frac{1}{32}$$ Answer: $$\boxed{\frac{1}{32}}$$ 25. **Problem:** $$\lim\limits_{(x,y,z) \to (1,3,4)} \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$$ Step 1: Substitute $x=1$, $y=3$, $z=4$. Step 2: Compute sum: $$1 + \frac{1}{3} + \frac{1}{4} = 1 + 0.3333 + 0.25 = 1.5833... = \frac{19}{12}$$ Answer: $$\boxed{\frac{19}{12}}$$