Algebra of limits

Theorem 3.1   If $${\lim_{x\to a} f(x)=L}$$ and $${\lim_{x\to a} g(x)=M}$$ then $${\lim_{x\to a} (f+g)(x)=L+M}$$.

PROOF:

Given $\epsilon$ we can use $${\lim_{x\to a} f(x)=L}$$ to find $\delta_f$ such that if $0<\vert x-a\vert<\delta_f$ then $\vert f(x)-L\vert<\epsilon/2$. We can use $${\lim_{x\to a} g(x)=M}$$ to find $\delta_g$ such that if $0<\vert x-a\vert<\delta_g$ then $\vert g(x)-M\vert<\epsilon/2$. Letting $\delta=\min(\delta_f,\delta_g)$ we get both, so that if $0<\vert x-a\vert<\delta$ then $\vert f(x)-L\vert<\epsilon/2$ and $\vert g(x)-M\vert<\epsilon/2$.Now by the triangle inequality

$$\begin{eqnarray*}\vert f(x)+g(x)-L-M\vert&=& \vert f(x)-L+g(x)-M\vert\\ &\leq&\v... ...rt+\vert g(x)-M\vert\\ &<& \epsilon/2 +\epsilon/2\\ &=& \epsilon\end{eqnarray*}$$

This proves $${\lim_{x\to a} (f+g)(x)=L+M}$$.

Commentary: We need to use the rules for finding $\delta$ from $\epsilon$ given by the two limits in the premises. The key idea is to split the allowed error $\epsilon$ in half and allocate half to $f$ and half to $g$.

Theorem 3.2   If $${\lim_{x\to a} f(x)=L}$$ and $${\lim_{x\to a} g(x)=M}$$ then $${\lim_{x\to a} f(x)g(x)=LM}$$.

PROOF:

If either $L=0$ or $M=0$ a special argument will be needed, though fortunately it is an easy one. We will prove the theorem for the case where both $L$ and $M$ are nonzero.

Given $\epsilon$ we will find $\delta$ by requiring several different conditions to be met.

Because $${\lim_{x\to a} g(x)=M}$$ we can find

$$\begin{eqnarray*} \delta_1 & \mbox{such that }& \mbox{if }0<\vert x-a\vert<\delt... ...\frac{\vert M\vert}2 < \vert g(x)\vert < \frac{3\vert M\vert}{2} \end{eqnarray*}$$

Because $${\lim_{x\to a} f(x)=L}$$ we can find $\delta_3$ such that if $0<\vert x-a\vert<\delta_3$ then $\vert f(x)-L\vert<\frac{\epsilon}{3\vert M\vert}$.

Let $\delta=\min(\delta_1,\delta_2,\delta_3)$. If $0<\vert x-a\vert<\delta$ then we get all three inequalities above. We seek a bound for $\vert f(x)g(x)-LM\vert$:

$$\begin{eqnarray*}\vert f(x)g(x)-LM\vert&=& \vert f(x)g(x)-Lg(x)+Lg(x)-LM\vert\\ ... ...\vert}\\ &=& \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{eqnarray*}$$

Commentary: This proof needed estimates on the size of $\vert g(x)\vert$ as well as keeping the values of $f(x)$ and $g(x)$ close enough to their respective limits. Adding and subtracting the same quantity so that the parts of the expression we wish to bound can be teased apart is an old mathematician's trick. The details of this proof are much more subtle than those of the proof for sums.

Theorem 3.3   If $${\lim_{x\to a} f(x)=L}$$ and $${\lim_{x\to a} g(x)=0}$$ then $${\lim_{x\to a} f(x)g(x)=0}$$.

PROOF:

Given an $\epsilon$ we can find $\delta_1$ so that if $0<\vert x-a\vert<\delta_1$ then $\vert f(x)-L\vert<\vert L\vert$ and a $\delta_2$ so that if $0<\vert x-a\vert<\delta_2$ then $\vert g(x)-0\vert<\frac{\epsilon}{2\vert L\vert}$. Let $\delta=\min(\delta_1,\delta_2)$. If $0<\vert x-a\vert<\delta$ then we get both $\vert f(x)-L\vert<\vert L\vert$ and $\vert g(x)-0\vert<\frac{\epsilon}{2\vert L\vert}$. The triangle inequality gives

$$\begin{eqnarray*}\vert f(x)g(x)-0\vert&=&\vert f(x)g(x)-Lg(x)+Lg(x)\vert\\ &\le... ...\vert L\vert}\\ &=& \frac{\epsilon}2+\frac{\epsilon}2 = \epsilon\end{eqnarray*}$$

as needed.

Theorem 3.4   If $${\lim_{x\to a} f(x)=L}$$ with $L\neq 0$ then $${\lim_{x\to a}\frac{1}{f(x)}=\frac1L}$$.

PROOF:

Given $\epsilon$ we can find

$$\begin{array}{lll} \delta_1 &\mbox{ such that if }0<\vert x-a... ...\vert f(x)-L\vert< \frac{\vert L\vert^2}{2}\epsilon \end{array}$$

Then let $\delta=\min(\delta_1,\delta_2)$ to get both of the inequalities. Then

$$\begin{eqnarray*}\left\vert\frac1{f(x)}-\frac1L\right\vert &=& \frac{\vert L-f(x... ...{2}\epsilon}{\frac{\vert L\vert}{2}\vert L\vert} \\ &=& \epsilon \end{eqnarray*}$$

as needed.

Corollary 3.5   If $${\lim_{x\to a} f(x)=L}$$ and $${\lim_{x\to a} g(x)=M\neq0}$$ then $${\lim_{x\to a} \frac{f(x)}{g(x)}=\frac{L}{M}}$$.

PROOF:

Write $${\frac{f(x)}{g(x)}}$$ as $${f(x)\frac1{g(x)}}$$ and then use the theorems on products and reciprocals.

Commentary: In the theorem on reciprocals we need to use the limit not being zero to bound the size of the denominator. Getting $\vert f(x)-L\vert<\frac{\vert L\vert}2$ forces $\frac{\vert L\vert}2 <\vert f(x)\vert<\frac{3\vert L\vert}2$. The rest is just decoration to make the final answer come out to $\epsilon$.