Equality behaves

Proposition 1.1   If $${\lim_{x\to a} f(x)=L}$$ and $${\lim_{x\to a} f(x)=M}$$ then $L=M$.

PROOF:

Suppose $L\neq M$, say $L>M$, then there must be a finite distance between $L$ and $M$. Let us see how this gets us into trouble: we let $\epsilon = (L-M)/2$ , then there must be $\delta_L$ such that if $0<\vert x-a\vert<\delta_f$ then $\vert f(x)-L\vert<(L-M)/2$ and $\delta_M$ such that if $0<\vert x-a\vert< \delta_M$ then $\vert f(x)-M\vert<(L-M)/2$. If we let $\delta=\min(\delta_M,\delta_L)$ then $0<\vert x-a\vert<\delta$ will guarantee both $\vert f(x)-M\vert<(L-M)/2$ and $\vert f(x)-L\vert<(L-M)/2$. Now the first of these guarantees that $f(x)<(L+M)/2$ and the second guarantees that $f(x) > (L+M)/2$ so we have a contradiction. Our only assumption was that $L\neq M$ so that must have been false, thus $L=M$.

Commentary: The confusion in this situation comes from the fact that the definition of a limit does not tell us that the $=$ in a limit statement has the same properties as the $=$ we use between numbers. In limits it says something about arbitrarily good approximation rather than something about specific values. This proposition tells us that the usual transitive law for $=$ also holds for limits. The proof itself is surprisingly indirect. We assume that the two limits are different and see how that gets us into trouble. The form of argument is ``to show $p \Rightarrow q$ assume $p$ and $\lnot q$ and argue to a contradiction, concluding $\lnot(p \wedge \lnot q)$, which is logically equivalent to $p \Rightarrow q$.

Other properties we associate with equality are easier:

Theorem 1.2   $${\lim_{x\to a} f(x)=L}$$ if and only if $${\lim_{x\to a} (f(x)-L)=0}$$.

PROOF:

Since $\vert f(x)-L\vert=\vert(f(x)-L)-0\vert$ finding a $\delta$ such that

$$\mbox{if } 0<\vert x-a\vert<\delta \mbox{ then }\vert f(x)-L\vert<\epsilon$$

is the same as finding a $\delta$ which makes

$$\vert(f(x)-L)-0\vert<\epsilon.$$