A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.2 Continuity

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.

   

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.

   

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

   

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.

    

    

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