MA005: Calculus I

Practice 1: $\quad f(x)=\frac{|x|}{x} \quad$ (Fig. 23) is continuous everywhere except at $\mathbf{x}=\mathbf{0}$ where this function is not defined.

If $a > 0$, then $\lim\limits{x \rightarrow a} \frac{|x|}{x}=1=\mathrm{f}(\mathrm{a})$ so $\mathrm{f}$ is continuous at $\mathrm{a}$. If $a < 0$, then $\lim\limits_{x \rightarrow a} \frac{|x|}{x}=-1=\mathrm{f}(\mathrm{a})$ so $\mathrm{f}$ is continuous at $\mathrm{a}$.

$\mathrm{f}(0)$ is not defined., $\lim\limits_{x \rightarrow 0^{-}} \frac{|x|}{x}=-1$ and $\lim\limits_{x \rightarrow 0^{+}} \frac{|x|}{x}=+1$ so $\lim\limits_{x \rightarrow 0} \frac{|x|}{x}$ does not exist.

Practice 2: (a) To prove that $\mathrm{kf}$ is continuous at $\mathrm{a}$, we need to prove that $\mathrm{kf}$ satisfies the definition of continuity at $\mathrm{a}$: $\lim\limits_{x \rightarrow a} \operatorname{kf}(\mathrm{x})=\mathrm{kf}(\mathrm{a})$.

Using results about limits, we know

(b) To prove that $\mathrm{f}-\mathrm{g}$ is continuous at $\mathrm{a}$, we need to prove that $\mathrm{f}-\mathrm{g}$ satisfies the definition of continuity at $\mathrm{a}$: $\lim\limits_{x \rightarrow a}(f(x)-g(x))=f(a)-g(a)$

Again using information about limits,

$\lim\limits_{x \rightarrow a}(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}))=\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})-\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{f}(\mathrm{a})-\mathrm{g}(\mathrm{a})$ (since $\mathrm{f}$ and $\mathrm{g}$ are both continuous at $\mathrm{a}$) so $\mathrm{f – g}$ is continuous at $\mathrm{a}$.