Algebra of limits: sums, scalar multiples, products and quotients

Reduction to null sequences: for any sequence $(z_k)$ of reals and any real $z$, the statements "$(z_k)$ converges to $z$" and "$(z_k-z)$ converges to $0$" are literally the same condition, because $\mathrm{\mid }(z_k-z)-0\mathrm{\mid }=\mathrm{\mid }{z}_k-z\mathrm{\mid }$ for every $k$.

Sum rule, in general form. Let $(u_k)\to u$ and $(v_k)\to v$ be any convergent sequences of reals and let $\epsilon >0$ be rational; take $K_1$ with $\mathrm{\mid }{u}_k-u\mathrm{\mid }<\epsilon \mathrm{/}2$ for $k\ge K_1$ and $K_2$ with $\mathrm{\mid }{v}_k-v\mathrm{\mid }<\epsilon \mathrm{/}2$ for $k\ge K_2$, and let $K$ be an index at least as large as both. For $k\ge K$, $\mathrm{\mid }(u_k+v_k)-(u+v)\mathrm{\mid }=\mathrm{\mid }(u_k-u)+(v_k-v)\mathrm{\mid }\le \mathrm{\mid }{u}_k-u\mathrm{\mid }+\mathrm{\mid }{v}_k-v\mathrm{\mid }<\epsilon \mathrm{/}2+\epsilon \mathrm{/}2=\epsilon$; hence $u_k+v_k\to u+v$, and in particular $x_k+y_k\to x+y$.

Boundedness: every convergent sequence of reals is bounded, and every constant sequence $(d)$ is bounded by $\mathrm{\mid }d\mathrm{\mid }$.

Quotient preparation. Assume $y\mathrm{\ne }0$ and $y_k\mathrm{\ne }0$ for every $k$. Then $\mathrm{\mid }y\mathrm{\mid }>0$ by [L2], so $\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2>0$; running the convergence test of $(y_k)$ with the real number $\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2$ as tolerance, which [L3] licenses, produces $K_0\in \mathbb{N}$ with $\mathrm{\mid }{y}_k-y\mathrm{\mid }<\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2$ for all $k\ge K_0$.

Scalar rule, in general form. Let $(u_k)\to u$ and let $d\in \mathbb{R}$. By step 1.1 the sequence $(u_k-u)$ is null and by step 1.3 the constant sequence $(d)$ is bounded, so $(d(u_k-u))=(d{u}_k-du)$ is null by [L5]; by step 1.1 again, $d{u}_k\to du$, and in particular $c{x}_k\to cx$.

Product rule, in general form. Let $(u_k)\to u$ and $(v_k)\to v$, and write $u_k{v}_k-uv=u_k(v_k-v)+v(u_k-u)$. By step 1.1 both $(v_k-v)$ and $(u_k-u)$ are null; by step 1.3 both $(u_k)$ and the constant sequence $(v)$ are bounded; so both $(u_k(v_k-v))$ and $(v(u_k-u))$ are null by [L5], and their sum is null by step 1.2 applied with both limits equal to $0$. By step 1.1, $u_k{v}_k\to uv$, and in particular $x_k{y}_k\to xy$.

Eventual lower bound. For every $k\ge K_0$, the reverse triangle inequality gives $\mathrm{\mid }{y}_k\mathrm{\mid }\ge \mathrm{\mid }y\mathrm{\mid }-\mathrm{\mid }y-y_k\mathrm{\mid }=\mathrm{\mid }y\mathrm{\mid }-\mathrm{\mid }{y}_k-y\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }-\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2=\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2>0$; so $\mathrm{\mid }{y}_k\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2$ for all $k\ge K_0$, and in particular $\mathrm{\mid }{y}_k\mathrm{\mid }\mathrm{\ne }0$ there.

Difference rule. Applying step 2.1 to the sequence $(y_k)$ with $d=-1$ gives $-y_k\to -y$; the sum rule of step 1.2 applied to $(x_k)$ and $(-y_k)$ then gives $x_k-y_k=x_k+(-y_k)\to x+(-y)=x-y$.

Reciprocal estimate. For $k\ge K_0$ we have $y_k\mathrm{\ne }0$ and $\mathrm{\mid }{y}_k\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2>0$, so [L7] applied to $0<\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2<\mathrm{\mid }{y}_k\mathrm{\mid }$ gives $0<\mathrm{\mid }{y}_k{\mathrm{\mid }}^{-1}<2\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$, and therefore $\mid {\displaystyle \frac{1}{y_k}}-{\displaystyle \frac{1}{y}}\mid ={\displaystyle \frac{\mathrm{\mid }y-y_k\mathrm{\mid }}{\mathrm{\mid }{y}_k\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}}\le {\displaystyle \frac{2}{\mathrm{\mid }y{\mathrm{\mid }}^2}}\mathrm{\mid }{y}_k-y\mathrm{\mid }$.

Reciprocal rule. Let $\epsilon >0$ be an arbitrary real and put $\eta :=\epsilon \mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}2$, a real $>0$; by [L3] there is $K_1$ with $\mathrm{\mid }{y}_k-y\mathrm{\mid }<\eta$ for all $k\ge K_1$. For every $k$ at least as large as both $K_0$ and $K_1$, step 3.2 gives $\mid {\displaystyle \frac{1}{y_k}}-{\displaystyle \frac{1}{y}}\mid \le {\displaystyle \frac{2}{\mathrm{\mid }y{\mathrm{\mid }}^2}}\mathrm{\mid }{y}_k-y\mathrm{\mid }<{\displaystyle \frac{2}{\mathrm{\mid }y{\mathrm{\mid }}^2}}\eta =\epsilon$; hence $1\mathrm{/}{y}_k\to 1\mathrm{/}y$.

Quotient rule. By step 4.1 the sequence $(1\mathrm{/}{y}_k)$ converges to $1\mathrm{/}y$, so the product rule of step 2.2 applied to $(x_k)$ and $(1\mathrm{/}{y}_k)$ gives $x_k\mathrm{/}{y}_k=x_k\cdot (1\mathrm{/}{y}_k)\to x\cdot (1\mathrm{/}y)=x\mathrm{/}y$.

All the claims are established: the sum rule in step 1.2, the scalar rule in step 2.1, the difference rule in step 3.1, the product rule in step 2.2, and the reciprocal and quotient rules in steps 4.1 and 5.1.