Calculus I · Getting Arbitrarily Close

Where this is heading

You now have the one tool Chapter 3 is built from. The slope of a curve at a point is the slope of the line joining two nearby points, $\dfrac{f(a+h)-f(a)}{h}$, in the limit as the two points slide together — that is, as $h\to0$. Substitute $h=0$ directly and you get $\tfrac00$, every single time. But you are no longer afraid of $\tfrac00$. You know it is not a wall; it is an invitation to cancel. Take that fear away and the derivative — the whole engine of the book — is just one more limit of exactly the kind you spent this chapter learning to read.

The slope of a curve is a limit. Turn the page and we’ll take it.

Problems

Six tiers, and they do different jobs. Fluency builds speed until the techniques are automatic — do all of them, fast. Practice is multi-step work where you must choose the right tool. Concept & counterexample has almost no computation; it tests whether you actually understand limits, and it is where most of the real learning hides — do not skip it. ◆ Diamond is Layer 2: proofs and stretch. Teach it asks you to explain, write up, or debug. Looking ahead quietly hands you Chapter 3. Several problems now exercise the chapter’s two theorems — the Intermediate and Extreme Value Theorems — alongside the limit techniques.

Fluency

1. By substitution: (a) $\displaystyle\lim_{x\to1}(x^3-2x+1)$; (b) $\displaystyle\lim_{x\to-2}(3x+1)$; (c) $\displaystyle\lim_{x\to0}\cos x$; (d) $\displaystyle\lim_{x\to9}\sqrt{x}$.
2. Factor and cancel: (a) $\displaystyle\lim_{x\to2}\frac{x^2-5x+6}{x-2}$; (b) $\displaystyle\lim_{x\to1}\frac{x^3-1}{x-1}$; (c) $\displaystyle\lim_{x\to3}\frac{x^2-x-6}{x^2-9}$.
3. Rationalize: (a) $\displaystyle\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}$; (b) $\displaystyle\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}$.
4. At infinity: (a) $\displaystyle\lim_{x\to\infty}\frac{3x^3-x}{2x^3+5}$; (b) $\displaystyle\lim_{x\to\infty}\frac{5x+1}{x^2+2}$; (c) $\displaystyle\lim_{x\to\infty}\frac{4x^2-3}{2x^2+x}$; (d) $\displaystyle\lim_{x\to\infty}\frac{2x^2+1}{x+1}$.
5. One-sided: for $f(x)=x$ if $x<1$ and $f(x)=x+2$ if $x\ge1$, find $\lim_{x\to1^-}f$, $\lim_{x\to1^+}f$, and say whether $\lim_{x\to1}f$ exists.
6. Continuity check (one line each): is $\tfrac1x$ continuous at $x=2$? at $x=0$? Is $\sqrt{x}$ continuous at $x=-1$?
7. Intermediate Value Theorem. Use the IVT to show each function has a root in the given interval (check the sign at each endpoint): (a) $x^3+x-1$ on $[0,1]$; (b) $\cos x - x$ on $[0,1]$; (c) $x^5-3x+1$ on $[0,1]$.
8. Extreme Value Theorem. Does the EVT guarantee a maximum and a minimum? Answer yes or no for each, with the reason: (a) $x^2$ on $[-1,3]$; (b) $\tfrac1x$ on $(0,1]$; (c) $x$ on $[0,5)$; (d) any continuous function on $[2,7]$.

Practice

1. $\displaystyle\lim_{x\to3}\frac{x^3-27}{x^2-9}$. (Difference of cubes on top; common factor with the bottom.)
2. Complex fraction: $\displaystyle\lim_{x\to3}\frac{\frac1x-\frac13}{x-3}$. (Combine the small fractions first, then cancel.)
3. An $\infty-\infty$ that hides a finite answer: $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+3x}-x\right)$. (Rationalize, then divide by the highest power.)
4. $\displaystyle\lim_{x\to\infty}\left(\sqrt{4x^2+x}-2x\right)$.
5. Find constants $a$ and $b$ that make $f$ continuous everywhere, where $f(x)=x^2$ for $x\le1$, $f(x)=ax+b$ for $1<x<3$, and $f(x)=4$ for $x\ge3$. (Two matching conditions, two unknowns.)
6. Is $f$ continuous at $x=1$, where $f(x)=\dfrac{x^2-1}{x-1}$ for $x<1$ and $f(x)=2x$ for $x\ge1$? Justify with the three-part continuity test. (A repaired hole meeting a line — do they actually agree?)
7. Bisection by hand. You showed in problem 7 that $x^3+x-1$ has a root in $[0,1]$. Do two steps of bisection: evaluate $f$ at the midpoint $0.5$, decide which half holds the root, then bisect that half. Report the interval of width $0.25$ that traps the root.
8. The IVT in the world. At 6 a.m. the temperature was $40°$F; at 2 p.m. it was $75°$F. Explain why, assuming temperature varies continuously, there must have been a moment when it read exactly $60°$F — and explain why this argument cannot tell you when.

Concept & counterexample

State true or false and justify — a one-line reason or, for a false claim, an explicit counterexample.

1. 1. If $\lim_{x\to a}f(x)=L$, then $f(a)=L$. (b) If $f(a)=5$, then $\lim_{x\to a}f(x)=5$. (c) If $\lim_{x\to a^-}f$ and $\lim_{x\to a^+}f$ both exist, then $\lim_{x\to a}f$ exists. (d) If $f$ and $g$ each fail to have a limit at $a$, then $f+g$ fails to have a limit at $a$.
2. Construct a function $f$ with $\lim_{x\to2}f(x)=4$ but $f(2)=10$. (Then say which type of discontinuity you built.)
3. Construct a single function that is continuous everywhere except for a jump at $x=0$ and a removable hole at $x=1$. (Write it piecewise; sketch it.)
4. Find the error. A student writes: “To find $\lim_{x\to3}\dfrac{x^2-9}{x-3}$, plug in $x=3$ to get $\tfrac00$. Since $\tfrac00$ is undefined, the limit does not exist.” Identify the exact mistake and give the correct answer.
5. Construct two functions $f$ and $g$, each with no limit as $x\to0$, whose product $fg$ does have a limit at $0$. (This is the multiplicative version of 17(d).)
6. Your friend says: “A function has a hole at $x=2$. Just define $f(2)$ to be any number you like and now it’s continuous.” When are they right, and when are they wrong? State the precise condition.
7. True or false, and justify: (a) If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have the same sign, then $f$ has no root in $[a,b]$. (b) A continuous function on an open interval $(a,b)$ always attains a maximum. (c) If a continuous $f$ takes both a positive and a negative value on $[a,b]$, it must have a root there. (d) The IVT tells you exactly how many roots a function has.
8. Construct a function on the closed interval $[0,1]$ that attains no maximum, and state precisely which hypothesis of the Extreme Value Theorem your example breaks. (Why must any such example be discontinuous?)

◆ Diamond — Layer 2 reach problems.

1. ◆ Prove $\displaystyle\lim_{x\to2}(3x-1)=5$ from the $\varepsilon$–$\delta$ definition, finding $\delta$ in terms of $\varepsilon$.
2. ◆ Prove $\displaystyle\lim_{x\to3}x^2 = 9$ from $\varepsilon$–$\delta$. (Hint: $|x^2-9|=|x-3|\,|x+3|$; first restrict $\delta\le1$ so $|x+3|\le7$, then take $\delta=\min(1,\tfrac{\varepsilon}{7})$.)
3. ◆ Prove $\displaystyle\lim_{x\to2}\frac1x=\frac12$ from $\varepsilon$–$\delta$. (The new wrinkle: the denominator. Show $\left|\tfrac1x-\tfrac12\right|=\tfrac{|x-2|}{2|x|}$, then restrict $\delta\le1$ to keep $|x|\ge1$ so the denominator can’t sabotage you; conclude $\delta=\min(1,2\varepsilon)$.)
4. ◆ Prove the sum law: if $\lim_{x\to a}f=L$ and $\lim_{x\to a}g=M$, then $\lim_{x\to a}(f+g)=L+M$. (Given $\varepsilon$, secure each piece within $\tfrac{\varepsilon}{2}$, then use the triangle inequality $|(f+g)-(L+M)|\le|f-L|+|g-M|$.)
5. ◆ Prove the squeeze-by-zero lemma: if $|f(x)|\le M$ for all $x$ near $a$ and $g(x)\to0$, then $f(x)g(x)\to0$. Then use it to re-derive $\lim_{x\to0}x^2\sin\tfrac1x=0$ in one line.
6. ◆ Prove the Squeeze Theorem for functions from the $\varepsilon$–$\delta$ definition.
7. ◆ Prove $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}=0$ using $\tfrac{\sin x}{x}\to1$. (Multiply by $\tfrac{1+\cos x}{1+\cos x}$, turn $1-\cos^2x$ into $\sin^2x$, then split.)
8. ◆ Using $\tfrac{\sin x}{x}\to1$, evaluate: (a) $\displaystyle\lim_{x\to0}\frac{\sin 5x}{x}$; (b) $\displaystyle\lim_{x\to0}\frac{\sin 2x}{\sin 3x}$; (c) $\displaystyle\lim_{x\to0}\frac{\tan x}{x}$. (For each, manufacture the $\tfrac{\sin(\text{stuff})}{\text{stuff}}$ shape by hand.)
9. ◆ A fixed point. Prove that every continuous function $f:[0,1]\to[0,1]$ has a fixed point — a $c$ with $f(c)=c$. (Apply the IVT to $g(x)=f(x)-x$: what are the signs of $g(0)$ and $g(1)$, given that $f$’s outputs lie in $[0,1]$?) This is a one-dimensional cousin of a celebrated theorem of Brouwer.

Teach it — explain it, write it up, or fix it for someone else

1. A classmate has absorbed the rule “if substitution gives $\tfrac00$, the limit doesn’t exist.” In three or four sentences written to them, explain why that’s wrong and give them the correct rule, using $\tfrac{x^2-9}{x-3}$ as your worked counterexample.
2. Write a complete, textbook-quality worked solution to $\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}$, narrating why you multiply by the conjugate and what it accomplishes, so a peer could reproduce the move on a new problem.
3. Debug your classmate. They claim $\displaystyle\lim_{x\to0}\frac{|x|}{x}=0$ “because the left side gives $-1$ and the right side gives $+1$, and those average to $0$.” Pinpoint the precise error and state what is actually true about this limit.
4. Design a problem. Invent a single piecewise function that has a removable hole at $x=-1$ and a jump at $x=2$. Write it out, then write the solution key that identifies each discontinuity and the one-sided limits at $x=2$.
5. In two or three sentences, teach the single most important idea of this chapter — that $\lim_{x\to a}f(x)$ and $f(a)$ are different questions — to someone who has never seen a limit, using one concrete example where the two answers disagree.
6. In three or four sentences with a real-world picture (a hiker’s altitude, a warming room, a car’s speedometer), teach a friend the Intermediate Value Theorem — why a continuously changing quantity cannot skip over a value on its way from one number to another — and name the single hypothesis that makes it true.

Looking ahead — you are about to compute derivatives with nothing but this chapter

1. For $f(x)=x^2$, expand and simplify $\dfrac{f(2+h)-f(2)}{h}$, then take $\lim_{h\to0}$. (You should get a clean number. That number is the slope of $x^2$ at $x=2$.)
2. Repeat 40 at a general point: find $\displaystyle\lim_{h\to0}\frac{(a+h)^2-a^2}{h}$ in terms of $a$. (You have just found the slope of $x^2$ at every point at once — keep the answer; Chapter 3 will call it $\tfrac{d}{dx}x^2$.)
3. For $f(x)=\tfrac1x$, find $\displaystyle\lim_{h\to0}\frac{\frac{1}{a+h}-\frac1a}{h}$. (This is problem 10’s complex-fraction move again — notice you already practiced exactly this.)
4. For $f(x)=\sqrt{x}$, find $\displaystyle\lim_{h\to0}\frac{\sqrt{a+h}-\sqrt{a}}{h}$. (Rationalize, as in problem 3.) When you finish 40–43, look back: every technique in this chapter existed to make these four limits computable. That is the whole point, and it is the first page of Chapter 3.
5. In problem 15 you trapped a root by bisection — reliable, but slow, halving the interval each step. Chapter 9 builds a far faster root-finder, Newton’s method, which uses the slope of $f$ (the derivative Chapter 3 is about) to leap toward the root. Predict one thing that could go wrong with a method that blindly follows the tangent line — a failure the plodding bisection method never suffers.

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