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Section 6.2 Continuity

These answers correspond to the problems in Section 1.3.

  1. \(c=\pi\text{.}\) Solve \(\displaystyle \lim _{x\to \pi ^-}f(x)=\lim _{x\to \pi ^-}f(x)\) for \(c\text{.}\) See Figure 6.2.1.

    Figure 6.2.1. \(c=\pi\)

  2. Let \(\displaystyle f(x)=2^x-\frac{10}{x}\text{.}\) Note that the domain of \(f\) is the set \(\mathbb{R}\backslash \{ 0\}\) and that on its domain, as a sum of two continuous function, \(f\) is continuous.

    1. Since \(f\) is continuous on \((0,\infty )\) and since \(\displaystyle \lim _{x\to 0^+}f(x)=-\infty\) and \(\displaystyle \lim _{x\to \infty}f(x)=\infty\) by the the Intermediate Value Property there is \(a\in (0,\infty )\) such that \(f(a)=0\text{.}\)

    2. For all \(x\in (-\infty ,0)\) we have that \(\displaystyle \frac{10}{x}\lt 0\) which implies that for all \(x\in (-\infty ,0)\) we have that all \(f(x)>0\text{.}\)

  3. See Figure 6.2.2. (a) (i) False; (ii)True; (b) (i) Yes; (ii) Yes; (c) (i) No; (ii) No.

    Figure 6.2.2. A piecewise defined function

  4. See Figure 6.2.3.

    1. Check that \(\displaystyle \lim _{x\to 1^-}f(x)=\lim _{x\to 1^+}f(x)=f(1)\text{.}\)

    2. \(\displaystyle \frac{1}{2}\text{.}\) Note \(\displaystyle \lim _{x\to 1^-}\frac{\frac{5+x}{2}-3}{x-1}=\frac{1}{2}\) and \(\displaystyle \lim _{x\to 1^+}\frac{(2+\sqrt{x})-3}{x-1}=\frac{1}{2}\text{.}\)

    Figure 6.2.3. A continuous function

  5. \(-1\text{.}\)

  6. \(\sqrt[3]{9}\text{.}\)

  7. \(\ds \frac{2}{3}\text{.}\)

  8. \(\ds -\frac{1}{\pi}\text{.}\)

  9. \(\displaystyle f(x)=\frac{x^2-9}{x-3}\) if \(x\not= 3\) and \(f(3)=0\text{.}\)

  10. See Figure 6.2.4.

  11. See Figure 6.2.4.

    Figure 6.2.4. Problems 6.2.10-11

  12. Consider the function \(g(x)=f(x)-x\text{.}\)