The Algebra of Limits

(Boundedness lemma.) Apply the definition of the limit with $\varepsilon = 1$. There exists $\delta_0 > 0$ such that for all $x \in D$ with $0 < |x - a| < \delta_0$,

By the triangle inequality,

Set $K = 1 + |L|$. This proves the lemma. $\blacksquare$

(1. Sum Law.) Technique: direct $\varepsilon$-$\delta$ verification, splitting the error via the triangle inequality. Let $\varepsilon > 0$. Since $\lim_{x\to a} f(x) = L$, applying the definition with tolerance $\varepsilon/2 > 0$ gives $\delta_1 > 0$ such that

Since $\lim_{x\to a} g(x) = M$, applying the definition with tolerance $\varepsilon/2 > 0$ gives $\delta_2 > 0$ such that

Let $\delta = \min(\delta_1, \delta_2) > 0$. If $x \in D$ and $0 < |x - a| < \delta$, then both conditions above hold simultaneously, so by the triangle inequality,

Since $\varepsilon > 0$ was arbitrary, $\lim_{x\to a}(f(x)+g(x)) = L+M$. $\blacksquare$

(2. Constant-Multiple Law.) Technique: direct $\varepsilon$-$\delta$ verification, treating $c=0$ separately. If $c = 0$, then $cf(x) = 0$ for all $x$, and the constant function $0$ trivially has limit $0 = c L$ at $a$ (given any $\varepsilon>0$, any $\delta>0$ works, since $|0 - 0| = 0 < \varepsilon$). Assume now $c \neq 0$. Let $\varepsilon > 0$. Then $\varepsilon / |c| > 0$, so applying the definition of $\lim_{x\to a} f(x) = L$ with tolerance $\varepsilon/|c|$, there is $\delta > 0$ such that

For such $x$,

Since $\varepsilon>0$ was arbitrary, $\lim_{x\to a} c f(x) = cL$. $\blacksquare$

(3. Product Law.) Technique: algebraic decomposition ("add and subtract a cross term") combined with the boundedness lemma. The key identity, valid for all $x \in D$, is

Indeed, expanding the right side:

which equals the left side since $ML = LM$. This is a purely algebraic identity, true for every $x$, with no limiting process invoked yet. By the boundedness lemma applied to $f$, there exist $\delta_0 > 0$ and $K > 0$ such that

Fix $\varepsilon > 0$. We will bound each term on the right of $(\ast)$ by $\varepsilon/2$. For the first term: since $K > 0$, choose (by the definition of $\lim_{x\to a} g(x) = M$, using tolerance $\dfrac{\varepsilon}{2K}>0$) a $\delta_1 > 0$ such that

Then for $0 < |x-a| < \min(\delta_0,\delta_1)$,

For the second term: if $M = 0$ this term is identically $0$ and trivially less than $\varepsilon/2$; if $M \ne 0$, choose (by the definition of $\lim_{x\to a} f(x)=L$, using tolerance $\dfrac{\varepsilon}{2|M|}>0$) a $\delta_2 > 0$ such that

so that for such $x$, $|M|\,|f(x)-L| < |M|\cdot \frac{\varepsilon}{2|M|} = \frac{\varepsilon}{2}$. (In the case $M=0$ we may simply set $\delta_2 = \delta_1$, since the bound $0 < \varepsilon/2$ holds trivially.) Now let $\delta = \min(\delta_0, \delta_1, \delta_2) > 0$. For $x \in D$ with $0 < |x - a| < \delta$, applying the triangle inequality to $(\ast)$:

Since $\varepsilon > 0$ was arbitrary, $\lim_{x\to a} f(x)g(x) = LM$. $\blacksquare$

(4. Quotient Law.) Technique: reduce to the special case $\lim_{x\to a} \frac{1}{g(x)} = \frac{1}{M}$ via the product law, after first establishing that $g$ is bounded away from $0$ near $a$. Step A: $g$ is bounded away from $0$ near $a$, and this gives the "there is a punctured neighborhood on which $g(x)\ne 0$" claim. Since $M \ne 0$, apply the definition of $\lim_{x\to a} g(x) = M$ with tolerance $\varepsilon = |M|/2 > 0$: there is $\delta_3 > 0$ such that

By the reverse triangle inequality, for such $x$,

In particular $g(x) \neq 0$ for all $x \in D$ with $0 < |x-a| < \delta_3$: this is exactly the punctured neighborhood claimed in the theorem, so $f(x)/g(x)$ is well-defined there. Moreover $|g(x)| > |M|/2$ on this set. Step B: $\lim_{x\to a} \dfrac{1}{g(x)} = \dfrac{1}{M}$. Let $\varepsilon > 0$. For $x \in D$ with $0 < |x-a| < \delta_3$,

Since $\lim_{x\to a} g(x) = M$, apply the definition with tolerance $\dfrac{\varepsilon |M|^2}{2} > 0$ to get $\delta_4 > 0$ such that

Let $\delta = \min(\delta_3, \delta_4) > 0$. For $x \in D$ with $0 < |x-a| < \delta$, combining the two displayed bounds,

Since $\varepsilon>0$ was arbitrary, $\lim_{x\to a} \frac{1}{g(x)} = \frac{1}{M}$. Step C: conclude via the Product Law. We have $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a}\frac{1}{g(x)} = \frac{1}{M}$, both already established (the second in Step B, which used only Step A and the basic definition, not the Product Law itself, so there is no circularity). By part (3) of this very theorem (Product Law), applied to the functions $f$ and $1/g$ on the common punctured neighborhood where both are defined,

This completes the proof of the Quotient Law, and of the theorem. $\blacksquare$

Write $f(x) - g(x) = f(x) + (-1)g(x)$. By the Constant-Multiple Law with $c=-1$, $\lim_{x\to a}(-1)g(x) = -M$. By the Sum Law applied to $f$ and $(-1)g$, $\lim_{x\to a}\big(f(x) + (-1)g(x)\big) = L + (-M) = L - M$. $\blacksquare$