MA005: Calculus I

The continuity of a function is defined in terms of limits, and all of these results about simple combinations of continuous functions follow from the results about simple combinations of limits in the Main Limit Theorem. Our hypothesis is that $\mathrm{f}$ and $\mathrm{g}$ are both continuous at a, so we can assume that $\lim\limits_{x \rightarrow a} f(x)=f(a)$ and $\quad \lim\limits_{x \rightarrow a} g(x)=g(a)$ and then use the appropriate part of the Main Limit Theorem. For example, $\lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}=\left\{\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})\right\}+\left\{\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})\right\}=\mathrm{f}(\mathrm{a})+\mathrm{g}(\mathrm{a})$, so $\mathrm{f}+\mathrm{g}$ is continuous at $\mathrm{a}$.

Practice 2: Prove: If $\mathrm{f}$ and $\mathrm{g}$ are continuous at $\mathrm{a}$, then $\mathrm{kf}$ and $\mathrm{f}-\mathrm{g}$ are continuous at $\mathrm{a}$. (\$\mathrm{k}$ a constant.)

This result will not be proved here, but the proof just formalizes the following line of reasoning:

The hypothesis that "$\mathrm{g}$ is continuous at $\mathrm{a}$" means that if $x$ is close to $\mathrm{a}$ then $g(x)$ will be close to $\mathrm{g}(\mathrm{a})$. Similarly, "$\mathrm{f}$ is continuous at $\mathrm{g} (\mathrm{a})$" means that if $\mathrm{g}(\mathrm{x})$ is close to $\mathrm{g}(\mathrm{a})$ then $f(g(x))=f \circ g(x)$ will be close to $f(g(a))=f \circ g(a)$. Finally, we can conclude that if $x$ is close to $a$ then $\mathrm{g}(\mathrm{x})$ is close to $\mathrm{g}(\mathrm{a})$ so $\mathrm{f}\circ\mathrm{g(x)}$  is close to $\mathrm{f}\circ\mathrm{g(a)}$ , and therefore $\mathrm{f}\circ\mathrm{g}$ is continuous at $\mathrm{x}=\mathrm{a}$.

The next theorem presents an alternate version of the limit condition for continuity, and we will use this alternate version occasionally in the future.

Theorem: $\quad \lim\limits_{x \rightarrow a} f(x)=f(a)$ if and only if $\quad \lim\limits_{h \rightarrow 0} f(a+h)=f(a)$.

Proof: Let's define a new variable $\mathrm{h}$ by $\mathrm{h}=\mathrm{x}-\mathrm{a}$ so $\mathrm{x}=\mathrm{a}+\mathrm{h}$

(Fig. 5). Then $x \rightarrow a$ if and only if $h=x-a \rightarrow 0$, so $\lim\limits_{x \rightarrow a} f(x)=\lim\limits_{h \rightarrow 0} f(a+h)$, and $\lim\limits_{x \rightarrow a} f(x)=f(a)$ if and only if $\lim\limits_{h \rightarrow 0} f(a+h)=f(a)$.

A function $\mathbf{f}$ is continuous at $\mathrm{a}$ if and only if $\lim\limits_{h \rightarrow 0} \mathbf{f}(\mathbf{a}+\mathbf{h})=\mathbf{f}(\mathbf{a})$.