Calculus/Proofs of Some Basic Limit Rules

Now that we have the formal definition of a limit, we can set about proving some of the properties we stated earlier in this chapter about limits.

We need to find a $$\delta>0$$ such that for every $$\varepsilon>0$$, $$|b-b|<\varepsilon$$ whenever $$0<|x-a|<\delta$$. $$|b-b|=0$$ and $$\varepsilon>0$$, so $$|b-b|<\varepsilon$$ is satisfied independent of any value of $$\delta$$ ; that is, we can choose any $$\delta$$ we like and the $$\varepsilon$$ condition holds.
 * Proof of the Constant Rule for Limits:

To prove that $$\lim_{x\to a}x=a$$, we need to find a $$\delta>0$$ such that for every $$\varepsilon>0$$ , $$|x-a|<\varepsilon$$ whenever $$0<|x-a|<\delta$$. Choosing $$\delta=\varepsilon$$ satisfies this condition.
 * Proof

Given the limit above, there exists in particular a $$\delta>0$$ such that $$\Big|f(x)-L\Big|<\frac{\varepsilon}{k}$$ whenever $$0<|x-a|<\delta$$, for some $$k>0$$ such that $$|c|0$$ , $$\Big|f(x)-L\Big|<\varepsilon$$ whenever $$|x-c|<\delta_f(\varepsilon)$$ , and $$\Big|g(x)-M\Big|<\varepsilon$$ whenever $$|x-c|<\delta_{g}(\varepsilon)$$. Adding the two inequalities gives $$\Big|f(x)-L\Big|+\big|g(x)-M\big|<2\varepsilon$$. By the triangle inequality we have $$\bigg|(f(x)-L)+(g(x)-M)\bigg|=\bigg|(f(x)+g(x))-(L+M)\bigg|\le\Big|f(x)-L\Big|+\Big|g(x)-M\Big|$$, so we have $$\bigg|(f(x)+g(x))-(L+M)\bigg|<2\varepsilon$$ whenever $$|x-c|<\delta_f(\varepsilon)$$ and $$|x-c|<\delta_{g}(\varepsilon)$$. Let $$\delta_{fg}(\varepsilon)$$ be the smaller of $$\delta_f(\tfrac{\varepsilon}{2})$$ and $$\delta_g(\tfrac{\varepsilon}{2})$$. Then this $$\delta$$ satisfies the definition of a limit for $$\lim_{x\to c}\Big[f(x)+g(x)\Big]$$ having limit $$L+M$$.
 * Proof

Define $$h(x)=-g(x)$$. By the Scalar Product Rule for Limits, $$\lim_{x\to c}h(x)=-M$$. Then by the Sum Rule for Limits, $$\lim_{x\to c}\Big[f(x)-g(x)\Big]=\lim_{x\to c}\Big[f(x)+h(x)\Big]=L-M$$.
 * Proof

Let $$\varepsilon$$ be any positive number. The assumptions imply the existence of the positive numbers $$\delta_1,\delta_2,\delta_3$$ such that
 * Proof
 * $$(1)\qquad\Big|f(x)-L\Big|<\frac{\varepsilon}{2(1+|M|)}$$ when $$0<|x-c|<\delta_1$$
 * $$(2)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon}{2(1+|L|)}$$ when $$0<|x-c|<\delta_2$$
 * $$(3)\qquad\Big|g(x)-M\Big|<1$$ when $$0<|x-c|<\delta_3$$

According to the condition (3) we see that
 * $$\Big|g(x)\Big|=\bigg|g(x)-M+M\bigg|\le\Big|g(x)-M\Big|+|M|<1+|M|$$ when $$0<|x-c|<\delta_3$$

Supposing then that $$0<|x-c|<\min\{\delta_1,\delta_2,\delta_3\}$$ and using (1) and (2) we obtain
 * $$\begin{align}\bigg|f(x)g(x)-LM\bigg|

&=\bigg|f(x)g(x)-Lg(x)+Lg(x)-LM\bigg|\\ &\le\bigg|f(x)g(x)-Lg(x)\bigg|+\bigg|Lg(x)-LM\bigg|\\ &=\Big|g(x)\Big|\cdot\Big|f(x)-L\Big|+|L|\cdot\Big|g(x)-M\Big|\\ &<(1+|M|)\frac{\varepsilon}{2(1+|M|)}+(1+|L|)\frac{\varepsilon}{2(1+|L|)}\\ &=\varepsilon \end{align}$$

If we can show that $$\lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}$$, then we can define a function, $$h(x)$$ as $$h(x)=\frac{1}{g(x)}$$ and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that $$\lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}$$.
 * Proof

Let $$\varepsilon$$ be any positive number. The assumptions imply the existence of the positive numbers $$\delta_1,\delta_2$$ such that
 * $$(1)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon|M|^2}{2}$$ when $$0<|x-c|<\delta_1$$
 * $$(2)\qquad\Big|g(x)-M\Big|<\frac{|M|}{2}$$ when $$0<|x-c|<\delta_{2}$$

According to the condition (2) we see that
 * $$|M|=|M-g(x)+g(x)|\le|M-g(x)|+|g(x)|<\frac{|M|}{2}+|g(x)| $$ so $$ |g(x)|>\frac{|M|}{2} $$ when $$0<|x-c|<\delta_2$$

which implies that
 * $$(3)\qquad\left|\frac{1}{g(x)}\right|<\frac{2}{|M|}$$ when $$0<|x-c|<\delta_2$$

Supposing then that $$0<|x-c|<\min\{\delta_1,\delta_2\}$$ and using (1) and (3) we obtain
 * $$\begin{align}\left|\frac{1}{g(x)}-\frac{1}{M}\right|&=\left|\frac{M-g(x)}{Mg(x)}\right|\\

&=\left|\frac{g(x)-M}{Mg(x)}\right|\\ &=\left|\frac{1}{g(x)}\right|\cdot\left|\frac{1}{M}\right|\cdot|g(x)-M|\\ &<\frac{2}{|M|}\cdot\frac{1}{|M|}\cdot|g(x)-M|\\ &<\frac{2}{|M|}\cdot\frac{1}{|M|}\cdot\frac{\varepsilon|M|^2}{2}\\ &=\varepsilon \end{align}$$

From the assumptions, we know that there exists a $$\delta$$ such that $$\Big|g(x)-L\Big|<\varepsilon$$ and $$\Big|h(x)-L\Big|<\varepsilon$$ when $$0<|x-c|<\delta$$. These inequalities are equivalent to $$L-\varepsilon<g(x)<L+\varepsilon$$ and $$L-\varepsilon<h(x)<L+\varepsilon$$ when $$0<|x-c|<\delta$$. Using what we know about the relative ordering of $$f(x),g(x)$$, and $$h(x)$$ , we have $$L-\varepsilon<g(x)\le f(x)\le h(x)<L+\varepsilon$$ when $$0<|x-c|<\delta$$. Then $$-\varepsilon<f(x)-L<\varepsilon$$ when $$0<|x-c|<\delta$$. So $$\Big|f(x)-L\Big|<\varepsilon$$ when $$0<|x-c|<\delta$$.
 * Proof