1. Problem 23(a): Prove $f(x)=3x-5$ is continuous at $x=2$. 1.1. Definition: $f$ is continuous at $a$ if for every $\epsilon>0$ there exists $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$. 1.2. Compute $f(2)=3\cdot 2-5=1$. 1.3. Compute the difference and factor: $|f(x)-f(2)|=|3x-5-1|=|3(x-2)|=3|x-2|$. 1.4. When dividing by $|x-2|$ show the cancellation: $$\frac{|f(x)-f(2)|}{|x-2|}=\frac{|3(x-2)|}{|x-2|}=|3|\frac{\cancel{|x-2|}}{\cancel{|x-2|}}=3$$ 1.5. Given $\epsilon>0$ choose $\delta=\epsilon/3$. 1.6. Then if $0<|x-2|<\delta$ we get $|f(x)-f(2)|=3|x-2|<3\delta=\epsilon$. 1.7. Therefore $f$ is continuous at $x=2$. 2. Problem 23(b): Prove $f(x)=x^{2}$ is continuous at $x=3$. 2.1. Definition: same as above. 2.2. Compute $f(3)=9$ and factor the difference: $|f(x)-f(3)|=|x^{2}-9|=|x-3||x+3|$. 2.3. Divide and cancel the $|x-3|$ factor: $$\frac{|f(x)-f(3)|}{|x-3|}=\frac{|x^{2}-9|}{|x-3|}=|x+3|\frac{\cancel{|x-3|}}{\cancel{|x-3|}}=|x+3|$$ 2.4. To control $|x+3|$ pick $\delta\le 1$ so that $|x-3|<1$ implies $2< x <4$ and hence $|x+3|\le 7$. 2.5. Given $\epsilon>0$ choose $\delta=\min(1,\epsilon/7)$. 2.6. Then if $0<|x-3|<\delta$ we have $|f(x)-f(3)|=|x-3||x+3|<\delta\cdot 7\le\epsilon$. 2.7. Therefore $x^{2}$ is continuous at $x=3$. 3. Problem 23(c): Prove $f(x)=1/x$ is continuous at $x=1/2$. 3.1. Compute $f(1/2)=2$ and simplify the difference: $|f(x)-2|=\left|\frac{1}{x}-2\right|=\frac{|1-2x|}{|x|}=\frac{2|x-1/2|}{|x|}$. 3.2. Divide by $|x-1/2|$ and cancel: $$\frac{|f(x)-2|}{|x-1/2|}=\frac{2|x-1/2|}{|x|}\frac{\cancel{|x-1/2|}}{\cancel{|x-1/2|}}=\frac{2}{|x|}$$ 3.3. Choose $\delta\le 1/4$ so that $|x-1/2|<\delta$ implies $x\in(1/4,3/4)$ and hence $|x|\ge 1/4$. 3.4. Given $\epsilon>0$ choose $\delta=\min(1/4,\epsilon/8)$. 3.5. Then if $0<|x-1/2|<\delta$ we have $|f(x)-2|\le 8|x-1/2|<8\delta\le\epsilon$. 3.6. Therefore $1/x$ is continuous at $x=1/2$. 4. Problem 24(a): Is $f(x)=\ln x$ uniformly continuous on $(0,1)$? 4.1. Claim: $\ln x$ is not uniformly continuous on $(0,1)$. 4.2. Counterexample using sequences: take $x_{n}=1/n$ and $y_{n}=1/(2n)$. 4.3. Then $|x_{n}-y_{n}|=1/(2n)\to 0$ as $n\to\infty$. 4.4. But $|\ln x_{n}-\ln y_{n}|=|\ln(1/n)-\ln(1/(2n))|=|\ln 2|$, a fixed positive number. 4.5. Hence there exists $\epsilon=|\ln 2|>0$ such that arbitrarily small differences in $x$ produce differences at least $\epsilon$, so $\ln x$ is not uniformly continuous on $(0,1)$. 5. Problem 24(b): Is $f(x)=x\sin x$ uniformly continuous on $[0,\infty)$? 5.1. Claim: $x\sin x$ is not uniformly continuous on $[0,\infty)$. 5.2. Take $x_{n}=n\pi$ and $y_{n}=n\pi+1/n$ for $n\in\mathbb{N}$. 5.3. Then $|x_{n}-y_{n}|=1/n\to 0$. 5.4. But $f(x_{n})=n\pi\sin(n\pi)=0$ and $f(y_{n})=(n\pi+1/n)\sin(1/n)\to \pi$ as $n\to\infty$ because $\sin(1/n)\sim 1/n$. 5.5. Thus $|f(x_{n})-f(y_{n})|\to \pi>0$, so $x\sin x$ is not uniformly continuous on $[0,\infty)$. 6. Problem 24(c): Is $f(x)=\sqrt{x}$ uniformly continuous on $[0,\infty)$? 6.1. Claim: $\sqrt{x}$ is uniformly continuous on $[0,\infty)$. 6.2. Use the inequality $|\sqrt{x}-\sqrt{y}|^{2}=x+y-2\sqrt{xy}=|x-y|+2(\min(x,y)-\sqrt{xy})\le |x-y|$, since $\sqrt{xy}\ge \min(x,y)$ for $x,y\ge0$. 6.3. Hence $|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}$ for all $x,y\ge0$. 6.4. Given $\epsilon>0$ choose $\delta=\epsilon^{2}$. 6.5. Then $|x-y|<\delta$ implies $|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}<\epsilon$. 6.6. Therefore $\sqrt{x}$ is uniformly continuous on $[0,\infty)$. 7. Problem 24(d): Is $f(x)=1/(x^{2}+1)$ uniformly continuous on $\mathbb{R}$? 7.1. Claim: $1/(x^{2}+1)$ is uniformly continuous on $\mathbb{R}$. 7.2. Compute the derivative $f'(x)=-\dfrac{2x}{(x^{2}+1)^{2}}$ and note $|f'(x)|=\dfrac{2|x|}{(x^{2}+1)^{2}}$. 7.3. The continuous function $|f'|$ attains a global maximum on $\mathbb{R}$, and that maximum is finite (one checks it occurs at $|x|=1/\sqrt{3}$), so there exists $M>0$ with $|f'(x)|\le M$ for all $x$. 7.4. By the Mean Value Theorem $|f(x)-f(y)|\le M|x-y|$ for all $x,y$, so $f$ is Lipschitz and hence uniformly continuous. 8. Problem 24(e): Is $f(x)=e^{x}$ uniformly continuous on $[0,\infty)$? 8.1. Claim: $e^{x}$ is not uniformly continuous on $[0,\infty)$. 8.2. Take $x_{n}=n$ and $y_{n}=n+1/n$. 8.3. Then $|x_{n}-y_{n}|=1/n\to 0$. 8.4. But $|e^{x_{n}}-e^{y_{n}}|=e^{n}|1-e^{1/n}|\sim e^{n}/n\to\infty$, so differences do not go to $0$. 8.5. Thus $e^{x}$ is not uniformly continuous on $[0,\infty)$. 9. Problem 25: Let continuous $f,g:[0,1]\to[0,\infty)$ satisfy $\sup_{[0,1]}f=\sup_{[0,1]}g$. Prove there exists $x_{0}\in[0,1]$ with $f(x_{0})=g(x_{0})$. 9.1. Since $f$ and $g$ are continuous on the compact interval $[0,1]$, each attains its maximum value, say $M_{f}=f(x_{f})$ and $M_{g}=g(x_{g})$. 9.2. Hypothesis gives $M_{f}=M_{g}=M$. 9.3. If $f(x)\ne g(x)$ for all $x$, then either $f(x)>g(x)$ for all $x$ or $f(x)g(x)$ for all $x$ then $M_{f}=f(x_{f})>g(x_{f})\le M_{g}$, contradiction with $M_{f}=M_{g}$. 9.5. Similarly for the other strict inequality. Thus there must exist $x_{0}$ with $f(x_{0})=g(x_{0})$. 10. Problem 26: Let $f:[0,1]\to[0,1]$ be continuous. Prove $f$ has a fixed point. 10.1. Define $g(x)=f(x)-x$ which is continuous on $[0,1]$. 10.2. Note $g(0)=f(0)-0\ge 0$ because $f(0)\in[0,1]$, and $g(1)=f(1)-1\le 0$ because $f(1)\in[0,1]$. 10.3. By the Intermediate Value Theorem there exists $x_{0}\in[0,1]$ with $g(x_{0})=0$, i.e. $f(x_{0})=x_{0}$. 11. Problem 27: Show $f(x)=x^{2}$ is uniformly continuous on $[0,1]$ but not on $\mathbb{R}$. 11.1. On $[0,1]$ the derivative $2x$ is bounded by $2$, so by the Mean Value Theorem $|x^{2}-y^{2}|\le 2|x-y|$ and $x^{2}$ is Lipschitz and uniformly continuous. 11.2. On $\mathbb{R}$ it is not uniformly continuous: take $x_{n}=n$ and $y_{n}=n+1/n$ then $|x_{n}-y_{n}|=1/n\to 0$ but $|x_{n}^{2}-y_{n}^{2}|=2+1/n^{2}\to 2\ne 0$, contradicting uniform continuity. 12. Problem 28: Prove that if $I$ is closed and bounded and $f:I\to\mathbb{R}$ is continuous then $f$ is uniformly continuous on $I$. 12.1. This is the Heine--Cantor theorem. 12.2. Sketch proof: given $\epsilon>0$ for each $x\in I$ choose $\delta_{x}>0$ such that $|f(y)-f(x)|<\epsilon$ whenever $|y-x|<\delta_{x}$, forming an open cover of the compact set $I$. 12.3. Extract a finite subcover and take $\delta$ as the minimum of finitely many positive radii; this $\delta$ works for all $x,y\in I$, proving uniform continuity. 13. Problem 29: Characterize non-uniform continuity by sequences. 13.1. Claim: $f$ is not uniformly continuous on $\mathbb{R}$ iff there exists $\epsilon>0$ and sequences $(x_{n})$, $(y_{n})$ with $|x_{n}-y_{n}|<1/n$ and $|f(x_{n})-f(y_{n})|\ge \epsilon$ for all $n$. 13.2. Proof forward: if $f$ is not uniformly continuous then some $\epsilon>0$ fails the definition, so for each $\delta=1/n$ there exist $x_{n},y_{n}$ with $|x_{n}-y_{n}|<1/n$ and $|f(x_{n})-f(y_{n})|\ge\epsilon$. 13.3. Proof backward: if such sequences exist then for that $\epsilon$ no $\delta$ works (take $\delta>0$ and choose $n>1/\delta$), so $f$ is not uniformly continuous. 14. Problem 30: Let $f(x)=x\sin(1/x)$ for $x\ne0$ and $f(0)=0$. Prove $f$ is continuous at $0$. 14.1. Note $|x\sin(1/x)|\le |x|$ for $x\ne0$ because $|\sin|\le 1$. 14.2. Given $\epsilon>0$ choose $\delta=\epsilon$. 14.3. Then if $|x|<\delta$ we have $|f(x)-f(0)|=|x\sin(1/x)|\le |x|<\epsilon$. 14.4. Hence $f$ is continuous at $0$. 15. Problem 31: Equivalence of continuity and sequential continuity. 15.1. (a) $\Rightarrow$ (b): If $f$ is continuous at $c$ and $x_{n}\to c$, then for every $\epsilon>0$ there is $\delta>0$ with $|x-c|<\delta$ implying $|f(x)-f(c)|<\epsilon$. For large $n$ we have $|x_{n}-c|<\delta$, so $f(x_{n})\to f(c)$. 15.2. (b) $\Rightarrow$ (a): Contrapositive: if $f$ is not continuous at $c$ then there exists $\epsilon>0$ such that for every $\delta>0$ there is $x$ with $|x-c|<\delta$ but $|f(x)-f(c)|\ge\epsilon$. Take $\delta=1/n$ to build a sequence $x_{n}\to c$ with $|f(x_{n})-f(c)|\ge\epsilon$, so $f(x_{n})\not\to f(c)$. Thus sequential continuity implies continuity. 16. Problem 32: Let $f(x)=1/x$ for $x\ne0$ and $f(0)=c$. Show $f$ is not continuous at $0$. 16.1. Take $x_{n}=1/n\to 0$. 16.2. Then $f(x_{n})=n\to\infty$, which cannot converge to the finite number $c$. 16.3. Hence $f$ is not continuous at $0$. 17. Problem 33: Lipschitz functions examples and continuity. 17.1. Two examples: $f_{1}(x)=ax+b$ with Lipschitz constant $|a|$, and $f_{2}(x)=\sin x$ which is Lipschitz with constant $1$ because $|\sin x-\sin y|\le |x-y|$ for all $x,y$. 17.2. Proof that Lipschitz implies continuous: if $|f(x)-f(y)|\le \alpha|x-y|$ for all $x,y$ and given $\epsilon>0$ choose $\delta=\epsilon/\alpha$. 17.3. Then $|x-y|<\delta$ implies $|f(x)-f(y)|\le \alpha|x-y|<\alpha\delta=\epsilon$. 17.4. Thus every Lipschitz function is uniformly continuous and hence continuous. Final answers summarized: 23(a) continuous at 2; 23(b) continuous at 3; 23(c) continuous at 1/2; 24: (a) not uniformly continuous, (b) not uniformly continuous, (c) uniformly continuous, (d) uniformly continuous, (e) not uniformly continuous; 25 exists $x_{0}$ with $f(x_{0})=g(x_{0})$; 26 fixed point exists; 27 uniformly continuous on $[0,1]$ but not on $\mathbb{R}$; 28 continuous on closed bounded interval implies uniform continuity; 29 sequence characterization proven; 30 continuous at 0; 31 equivalence proven; 32 not continuous at 0; 33 examples and statement that Lipschitz implies continuous.