Meaning and Algebra of Limits

7.1.1 Limit of a function. Suppose x is a variable and a is a constant. If x takes values closer and closer to a but never actually equal to a, we say x tends to a, written $x\to a$. When x approaches from values larger than a (for example $a+\tfrac12,\ a+\tfrac14,\ a+\tfrac18,\dots$) we write $x\to a^{+}$; when it approaches from values smaller than a we write $x\to a^{-}$. For any polynomial $P(x)$, direct substitution always works: $\lim_{x\to a}P(x)=P(a)$. For a rational function $f(x)=P(x)/Q(x)$ there are three distinct possibilities as $x\to a$: (1) if $Q(a)\ne0$, direct substitution again works and $\lim_{x\to a}f(x)=P(a)/Q(a)$; (2) if $Q(a)=0$ and $P(a)=0$ then $(x-a)$ is a common factor of both, so writing $P(x)=(x-a)^rP_1(x)$ and $Q(x)=(x-a)^sQ_1(x)$, the factor cancels and if $r=s$ the limit is $P_1(a)/Q_1(a)$, if $r>s$ the limit is $0$, and if $r<s$ we fall into case (3); (3) if $Q(a)=0$ but $P(a)\ne0$, the limit does not exist (the function blows up). This three-way split is exactly what the Method of Factorization (7.2) and Method of Rationalization (7.3) automate.\n\n7.1.2 The precise (epsilon-delta) definition. Saying informally that 'f(x) gets close to l as x gets close to a' is made rigorous as follows: given any tolerance $\epsilon>0$ (however small), there must exist some window-width $\delta>0$ such that whenever x is within that window of a but not equal to a itself ($0<|x-a|<\delta$), the output f(x) is guaranteed to land within the $\epsilon$-tolerance of l ($|f(x)-l|<\epsilon$). If such a $\delta$ can always be found no matter how small $\epsilon$ is demanded, we declare $f(x)\to l$ as $x\to a$. Proving this in practice means starting from the inequality $|f(x)-l|<\epsilon$, simplifying it algebraically until it reads $|x-a|<(\text{something in }\epsilon)$, and then choosing $\delta$ to be that something (or smaller). Two worked patterns recur: for a straight line like $f(x)=3x+1$ the algebra is direct — $|(3x+1)-1|<\epsilon \iff |x|<\epsilon/3$, so $\delta=\epsilon/3$ works. For a curve like $f(x)=x^2$ (limit 9 at $a=3$) the factor $|x+3|$ appearing alongside $|x-3|$ has no fixed bound on its own, so a first restriction $\delta\le1$ is imposed to trap $|x+3|<7$, after which $\delta=\min\{\epsilon/7,\,1\}$ finishes the proof — the general lesson being that whenever a stray extra factor like $|x+3|$ shows up, cap $\delta$ at $1$ first to get a numeric bound on that factor, then solve for the rest.\n\n7.1.3 One-sided limits. $\lim_{x\to a^{-}}f(x)$ and $\lim_{x\to a^{+}}f(x)$, when they exist, are called the left-hand and right-hand (one-sided) limits.\n\n7.1.4 Left-hand limit. Formally: given $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-l|<\epsilon$ for every x with $a-\delta<x<a$ (approach strictly from below); then $\lim_{x\to a^{-}}f(x)=l$.\n\n7.1.5 Right-hand limit. Symmetrically: given $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-l|<\epsilon$ for every x with $a<x<a+\delta$ (approach strictly from above); then $\lim_{x\to a^{+}}f(x)=l$.\n\n7.1.6 Existence of a limit at x=a. The full (two-sided) limit exists, and equals l, exactly when both one-sided limits exist and agree: $\lim_{x\to a^{+}}f(x)=\lim_{x\to a^{-}}f(x)=l \implies \lim_{x\to a}f(x)=l$. If the two one-sided limits disagree, $\lim_{x\to a}f(x)$ simply does not exist — this is the standard test for a piecewise-defined function. Worked illustration: for $f(x)=[x]$ (the greatest-integer function) restricted to $2\le x\le4$, so that $[x]=2$ on $[2,3)$ and $[x]=3$ on $[3,4)$, the right-hand limit at $x=3$ is $3$ while the left-hand limit is $2$ — they disagree, so $\lim_{x\to3}[x]$ does not exist, even though the limit at a non-integer point such as $x=2.7$ exists trivially (equal to the constant value $2$ on both sides). A second illustration: for $f(x)=3x+1$ when $x<1$ and $f(x)=7x^2-3$ when $x\ge1$, the left-hand limit at $x=1$ is $\lim_{x\to1^-}(3x+1)=4$ and the right-hand limit is $\lim_{x\to1^+}(7x^2-3)=4$; since these agree, $\lim_{x\to1}f(x)=4$.\n\n7.1.7 Algebra of limits. If $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$, then: (1) $\lim_{x\to a}[f(x)\pm g(x)]=l\pm m$; (2) $\lim_{x\to a}[f(x)\times g(x)]=l\times m$; (3) $\lim_{x\to a}[k\,f(x)]=kl$ for any constant k; (4) $\lim_{x\to a}[f(x)/g(x)]=l/m$ provided $m\ne0$. Standing building blocks used everywhere: $\lim_{x\to a}k=k$; $\lim_{x\to a}x=a$; $\lim_{x\to a}x^n=a^n$; $\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$; and for any polynomial $p(x)$, $\lim_{x\to a}p(x)=p(a)$. A standing warning: before ever cancelling or substituting, always check whether the denominator's limit is zero — and if it is, whether the numerator's limit is also zero (0/0, needing factorization or rationalization) as opposed to a non-zero numerator over a zero denominator (limit does not exist).\n\n7.1.8 The standard power-difference theorem. $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$ for $n\in\mathbb N$, $a>0$. Proof: using the factorization identity $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+b^{n-1})$ with the roles $x,a$ in place of $a,b$, the numerator factors as $(x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$; the $(x-a)$ cancels against the denominator (valid since $x\ne a$ on the way to the limit), leaving $\lim_{x\to a}(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$, which by direct substitution is $n$ copies of $a^{n-1}$ added together, i.e. $na^{n-1}$. An alternative proof substitutes $x-a=h$ (so $x=a+h$ and $h\to0$ as $x\to a$), expands $(a+h)^n$ by the binomial theorem, and simplifies — the same result falls out. The theorem extends beyond positive integers: for a negative integer $n=-m$, $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=-ma^{-m-1}$; and for a fractional exponent $n=p/q$ (with $q\ne0$), the same formula $na^{n-1}=\tfrac{p}{q}a^{p/q-1}$ still holds. This single formula is the engine behind almost every question in Exercise 7.1: worked examples include $\lim_{x\to5}\frac{x^4-625}{x-5}=4(5)^3=500$; $\lim_{x\to2}\frac{x^7-128}{x^5-32}=\frac{7(2)^6}{5(2)^4}=\frac{28}{5}$ (dividing one $(x^n-a^n)/(x-a)$ ratio by another); finding $n=3$ from $\lim_{x\to4}\frac{x^n-4^n}{x-4}=48$ by matching $n(4)^{n-1}=48=3(4)^2$; and $\lim_{x\to1}\frac{2x-2}{\sqrt[3]{26+x}-3}=54$, found by substituting $26+x=t^3$ so the cube root disappears and the ratio becomes a plain $(t^3-3^3)/(t-3)$ limit equal to $3(3)^2=27$, then doubling for the factor of 2 in the numerator.