Exercises — Epsilon-delta definition of a limit — formal proofs

$\forall \epsilon >0\exists \delta >0\text{such that}0<|x-a|<\delta ⟹|f(x)-L|<\epsilon .$ The $0<$ part excludes $x=a$ itself: a limit describes the approach to $a$, and must not care whether $f$ is even defined at $a$.

The adversary chooses $\epsilon$ first; you reply with $\delta$ second. Your $\delta$ may depend on $\epsilon$ (that is the whole point — a smaller ring demands a smaller wobble). Your $\delta$ may not depend on $x$, because $x$ is chosen after you commit to $\delta$.

The output distance is $|f(x)-L|=|7-7|=0$, which is $<\epsilon$ for every $\epsilon >0$, regardless of $x$. Proof. Let $\epsilon >0$. Choose $\delta =1$ (any positive number works). If $0<|x-5|<\delta$, then $|7-7|=0<\epsilon$. $■$

Scratchwork. Output distance: $|(3x-5)-7|=|3x-12|=3|x-4|.$ The factor is the constant $3$. To force $3|x-4|<\epsilon$ we need $|x-4|<\epsilon /3$. Take $\delta =\epsilon /3$. Proof. Let $\epsilon >0$, choose $\delta =\epsilon /3$. If $0<|x-4|<\delta$ then $|(3x-5)-7|=3|x-4|<3\delta =3\cdot \frac{\epsilon }{3}=\epsilon .■$

Scratchwork. $|(-4x+2)-6|=|-4x-4|=4|x+1|=4|x-(-1)|$. The absolute value swallows the minus sign; the factor is $4$. Take $\delta =\epsilon /4$. Proof. Let $\epsilon >0$, choose $\delta =\epsilon /4$. If $0<|x+1|<\delta$ then $|(-4x+2)-6|=4|x+1|<4\delta =\epsilon .■$

$\left|\frac{x}{2}-1\right|=\frac{1}{2}|x-2|$. The factor is $\frac{1}{2}$. To make $\frac{1}{2}|x-2|<\epsilon$ we need $|x-2|<2\epsilon$. Take $\delta =2\epsilon$. Proof. Let $\epsilon >0$, choose $\delta =2\epsilon$. If $0<|x-2|<\delta$ then $\left|\frac{x}{2}-1\right|=\frac{1}{2}|x-2|<\frac{1}{2}\delta =\epsilon .$ $■$ Note the factor was less than 1, so $\delta$ came out bigger than $\epsilon$ — a gentle function tolerates a wide window.

Scratchwork. $|x^2-9|=|x-3||x+3|$. The factor $|x+3|$ is not constant — bound it first. Impose the preliminary $|x-3|<1$, so $2<x<4$, hence $5<x+3<7$ and $|x+3|<7$. Then $|x^2-9|<7|x-3|.$ Need $7|x-3|<\epsilon$, i.e. $|x-3|<\epsilon /7$. Enforce both with $\boxed{\delta =\min (1,\epsilon /7)}$. Proof. Let $\epsilon >0$, $\delta =\min (1,\epsilon /7)$. If $0<|x-3|<\delta$: since $\delta \le 1$, $|x+3|<7$; since $\delta \le \epsilon /7$, $|x^2-9|=|x-3||x+3|<7|x-3|<7\cdot \frac{\epsilon }{7}=\epsilon .■$

Scratchwork. $x^2+x-2=(x-1)(x+2)$, so $|f(x)-L|=|x-1||x+2|$. Preliminary $|x-1|<1$ gives $0<x<2$, so $2<x+2<4$ and $|x+2|<4$. Then $|f(x)-2|<4|x-1|$. Need $|x-1|<\epsilon /4$. Take $\delta =\min (1,\epsilon /4)$. Proof. Let $\epsilon >0$, $\delta =\min (1,\epsilon /4)$. If $0<|x-1|<\delta$: $|x+2|<4$ and $|x^2+x-2|=|x-1||x+2|<4|x-1|<4\cdot \frac{\epsilon }{4}=\epsilon .■$

Scratchwork. $\left|\frac{1}{x}-\frac{1}{5}\right|=\left|\frac{5-x}{5x}\right|=\frac{|x-5|}{5|x|}.$ The dangerous piece is $\frac{1}{|x|}$: if $x$ could reach $0$ it would blow up. So we must keep $x$ away from $0$. Preliminary $|x-5|<1$ gives $4<x<6$. Since $x>4>0$ on this interval, $x$ is positive, so $|x|=x>4$; hence $\frac{1}{|x|}<\frac{1}{4}$. Then $\left|\frac{1}{x}-\frac{1}{5}\right|=\frac{|x-5|}{5|x|}<\frac{|x-5|}{5\cdot 4}=\frac{|x-5|}{20}.$ Need $\frac{|x-5|}{20}<\epsilon$, i.e. $|x-5|<20\epsilon$. Take $\delta =\min (1,20\epsilon )$. Proof. Let $\epsilon >0$, $\delta =\min (1,20\epsilon )$. If $0<|x-5|<\delta$: then $4<x<6$, so $x>0$ and $|x|=x>4$; therefore $\left|\frac{1}{x}-\frac{1}{5}\right|=\frac{|x-5|}{5|x|}<\frac{|x-5|}{20}<\frac{20\epsilon }{20}=\epsilon .■$

Key fact: $\left|\sin \theta \right|\le 1$ for every $\theta$, so $\left|\sin \frac{1}{x}\right|\le 1$ whenever $x\ne 0$. Thus $\left|x\sin \frac{1}{x}-0\right|=|x|\left|\sin \frac{1}{x}\right|\le |x|=|x-0|.$ The wild oscillating factor is bounded by the constant $1$ — no preliminary restriction needed. Take $\delta =\epsilon$. Proof. Let $\epsilon >0$, choose $\delta =\epsilon$. If $0<|x-0|<\delta$ then $\left|x\sin \frac{1}{x}\right|\le |x|<\delta =\epsilon .■$ This is the squeeze idea living inside an $\epsilon$–$\delta$ proof; see Limit Laws (sum, product, quotient).

Here $f(x)=1$ for $x>0$ and $f(x)=-1$ for $x<0$. Negation: show some $\epsilon >0$ defeats every $\delta$ and every candidate $L$. Take $\epsilon =1$. Fix any candidate $L$ and any $\delta >0$. Construct two explicit points inside the window: let $x_{+}=\delta /2>0$ and $x_{-}=-\delta /2<0$. Both satisfy $0<|x|<\delta$, and $f(x_{+})=1$, $f(x_{-})=-1$. If both were within $1$ of $L$, then by the triangle inequality $2=|1-(-1)|=|(1-L)+(L-(-1))|\le |1-L|+|L-(-1)|<1+1=2,$ a strict contradiction ($2<2$ is false). So at least one of $x_{+},x_{-}$ lies $\ge 1$ from $L$. Since we produced such points for every $\delta >0$ and every $L$, the limit does not exist. $■$ Why $\epsilon =1$? The two output values are $2$ apart; a band of half-width $1$ (full width $2$) cannot contain both.

For $x>0$, output distance is $|\sqrt{x}-0|=\sqrt{x}$. We want $\sqrt{x}<\epsilon$, which (squaring, valid since both sides $\ge 0$) is $x<{\epsilon }^2$. Take $\delta ={\epsilon }^2$. Proof. Let $\epsilon >0$, choose $\delta ={\epsilon }^2$. If $0<x<\delta ={\epsilon }^2$ then $\sqrt{x}<\sqrt{{\epsilon }^2}=\epsilon$. $■$ Note the non-linear link: to shrink the output by a factor, the input window must shrink by its square.

Idea: we must control the output distance $|(f+g)-(L+M)|$. Split it with the triangle inequality: $|(f(x)+g(x))-(L+M)|\le |f(x)-L|+|g(x)-M|.$ Each piece can be forced below $\epsilon /2$, so the sum stays below $\epsilon$. Splitting $\epsilon$ in half is the trick. Proof. Let $\epsilon >0$. Since $\lim f=L$, there is ${\delta }_1>0$ with $0<|x-a|<{\delta }_1\Rightarrow |f(x)-L|<\epsilon /2$. Since $\lim g=M$, there is ${\delta }_2>0$ with $0<|x-a|<{\delta }_2\Rightarrow |g(x)-M|<\epsilon /2$. Choose $\delta =\min ({\delta }_1,{\delta }_2)$. If $0<|x-a|<\delta$, then both conditions hold, so $|(f+g)-(L+M)|\le |f-L|+|g-M|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .■$ This is exactly the sum rule in Limit Laws (sum, product, quotient).

Scratchwork. Factor: $x^3-8=(x-2)(x^2+2x+4)$, so $|x^3-8|=|x-2||x^2+2x+4|$. Preliminary $|x-2|<1$ gives $1<x<3$. On this interval $x^2+2x+4$ is increasing and positive; at $x=3$ it equals $9+6+4=19$, so $|x^2+2x+4|<19$. Then $|x^3-8|<19|x-2|$. Need $|x-2|<\epsilon /19$. Take $\delta =\min (1,\epsilon /19)$. Proof. Let $\epsilon >0$, $\delta =\min (1,\epsilon /19)$. If $0<|x-2|<\delta$: since $\delta \le 1$, $|x^2+2x+4|<19$; since $\delta \le \epsilon /19$, $|x^3-8|=|x-2||x^2+2x+4|<19|x-2|<19\cdot \frac{\epsilon }{19}=\epsilon .■$

Simplify first (for $h\ne 0$): $\frac{(3+h{)}^2-9}{h}=\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}=6+h.$ So the output distance is $|(6+h)-6|=|h|=|h-0|$. The factor is the constant $1$. Take $\delta =\epsilon$. Proof. Let $\epsilon >0$, choose $\delta =\epsilon$. If $0<|h-0|<\delta$, then the quotient equals $6+h$ (defined since $h\ne 0$), and $\left|\frac{(3+h{)}^2-9}{h}-6\right|=|h|<\delta =\epsilon .■$ Hence $f^{\prime }(3)=6$, matching $\frac{d}{dx}{x}^2=2x$ at $x=3$.