Limits and Continuity: Get Stronger Answer Key

The Limit Laws

In the following exercises (1-2), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

  1. Use constant multiple law and difference law: [latex]\underset{x\to 0}{\lim}(4x^2-2x+3)=4\underset{x\to 0}{\lim}x^2-2\underset{x\to 0}{\lim}x+\underset{x\to 0}{\lim}3=3[/latex]
  2. Use root law: [latex]\underset{x\to -2}{\lim}\sqrt{x^2-6x+3}=\sqrt{\underset{x\to -2}{\lim}(x^2-6x+3)}=\sqrt{19}[/latex]
  3. [latex]49[/latex]
  4. [latex]1[/latex]
  5. [latex]-\frac{5}{7}[/latex]
  6. [latex]\underset{x\to 4}{\lim}\frac{x^2-16}{x-4}=\frac{16-16}{4-4}=\frac{0}{0}[/latex]; then, [latex]\underset{x\to 4}{\lim}\frac{x^2-16}{x-4}=\underset{x\to 4}{\lim}\frac{(x+4)(x-4)}{x-4}=8[/latex]
  7. [latex]\underset{x\to 6}{\lim}\frac{3x-18}{2x-12}=\frac{18-18}{12-12}=\frac{0}{0}[/latex]; then, [latex]\underset{x\to 6}{\lim}\frac{3x-18}{2x-12}=\underset{x\to 6}{\lim}\frac{3(x-6)}{2(x-6)}=\frac{3}{2}[/latex]
  8. [latex]\underset{t \to 9}{\lim}\frac{t-9}{\sqrt{t}-3}=\frac{9-9}{3-3}=\frac{0}{0}[/latex]; then, [latex]\underset{t\to 9}{\lim}\frac{t-9}{\sqrt{t}-3}=\underset{t\to 9}{\lim}\frac{t-9}{\sqrt{t}-3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\underset{t\to 9}{\lim}(\sqrt{t}+3)=6[/latex]
  9. [latex]\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\tan \theta}=\frac{\sin \pi}{\tan \pi}=\frac{0}{0}[/latex]; then, [latex]\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\tan \theta}=\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}}=\underset{\theta \to \pi}{\lim}\cos \theta =-1[/latex].
  10. [latex]\underset{x\to 1/2}{\lim}\frac{2x^2+3x-2}{2x-1}=\frac{\frac{1}{2}+\frac{3}{2}-2}{1-1}=\frac{0}{0}[/latex]; then, [latex]\underset{x\to 1/2}{\lim}\frac{2x^2+3x-2}{2x-1}=\underset{x\to 1/2}{\lim}\frac{(2x-1)(x+2)}{2x-1}=\frac{5}{2}[/latex]
  11. [latex]\underset{x\to 6}{\lim}2f(x)g(x)=2\underset{x\to 6}{\lim}f(x)\underset{x\to 6}{\lim}g(x)=72[/latex]
  12. [latex]\underset{x\to 6}{\lim}(f(x)+\frac{1}{3}g(x))=\underset{x\to 6}{\lim}f(x)+\frac{1}{3}\underset{x\to 6}{\lim}g(x)=7[/latex]
  13. [latex]\underset{x\to 6}{\lim}\sqrt{g(x)-f(x)}=\sqrt{\underset{x\to 6}{\lim}g(x)-\underset{x\to 6}{\lim}f(x)}=\sqrt{5}[/latex]
  14. [latex]\underset{x\to 6}{\lim}[(x+1)\cdot f(x)]=(\underset{x\to 6}{\lim}(x+1))(\underset{x\to 6}{\lim}f(x))=28[/latex].
  15. The graph of a piecewise function with two segments. The first is the parabola x^2, which exists for x<=3. The vertex is at the origin, it opens upward, and there is a closed circle at the endpoint (3,9). The second segment is the line x+4, which is a linear function existing for x > 3. There is an open circle at (3, 7), and the slope is 1.
    1. [latex]9[/latex]
    2. [latex]7[/latex]
  16. The graph of a piecewise function with two segments. The first segment is the parabola x^2 – 2x + 1, for x < 2. It opens upward and has a vertex at (1,0). The second segment is the line 3-x for x>= 2. It has a slope of -1 and an x intercept at (3,0).
    1. [latex]1[/latex]
    2. [latex]1[/latex]
  17. [latex]\underset{x\to -3^-}{\lim}(f(x)-3g(x))=\underset{x\to -3^-}{\lim}f(x)-3\underset{x\to -3^-}{\lim}g(x)=0+6=6[/latex]
  18. [latex]\underset{x\to -5}{\lim}\frac{2+g(x)}{f(x)}=\frac{2+(\underset{x\to -5}{\lim}g(x))}{\underset{x\to -5}{\lim}f(x)}=\frac{2+0}{2}=1[/latex]
  19. [latex]\underset{x\to 1}{\lim}\sqrt[3]{f(x)-g(x)}=\sqrt[3]{\underset{x\to 1}{\lim}f(x)-\underset{x\to 1}{\lim}g(x)}=\sqrt[3]{2+5}=\sqrt[3]{7}[/latex]
  20. [latex]\underset{x\to -9}{\lim}(x\cdot f(x)+2g(x))=(\underset{x\to -9}{\lim}x)(\underset{x\to -9}{\lim}f(x))+2\underset{x\to -9}{\lim}(g(x))=(-9)(6)+2(4)=-46[/latex]

Continuity

For the following exercises (1-4), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

  1. The function is defined for all [latex]x[/latex] in the interval [latex](0,\infty)[/latex].
  2. Removable discontinuity at [latex]x=0[/latex]; infinite discontinuity at [latex]x=1[/latex]
  3. [latex]f(x)=\dfrac{5}{e^x-2}[/latex] Infinite discontinuity at [latex]x=\ln 2[/latex]
  4. [latex]H(x)= \tan 2x[/latex] Infinite discontinuities at [latex]x=\frac{(2k+1)\pi}{4}[/latex], for [latex]k=0, \, \pm 1, \, \pm 2, \, \pm 3, \cdots[/latex]
  5. No. It is a removable discontinuity.
  6. Yes. It is continuous.
  7. Yes. It is continuous.
  8. [latex]k=-5[/latex]
  9. [latex]k=-1[/latex]
  10. [latex]k=\frac{16}{3}[/latex]
  11. A graph of the given piecewise function containing two segments. The first, x^3, exists for x < 1 and ends with an open circle at (1,1). The second, 3x, exists for x > 1. It beings with an open circle at (1,3).

    It is not possible to redefine [latex]f(1)[/latex] since the discontinuity is a jump discontinuity.

  12. Answers may vary; see the following example:

    A graph of a piecewise function with several segments. The first is an increasing line that exists for x < -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 <= x < -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 < x <= 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x > 3.

  13. Answers may vary; see the following example:

    The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x < 1. It ends at (1,1). The second part is an increasing line that exists for x > 1. It begins at (1,3)./li>

  14. False. It is continuous over [latex](−\infty,0) \cup (0,\infty)[/latex].
  15. False. Consider [latex]f(x)=\begin{cases} x & \text{ if } \, x \ne 0 \\ 4 & \text{ if } \, x = 0 \end{cases}[/latex]
  16. False. Consider [latex]f(x)= \cos (x)[/latex] on [latex][-\pi, 2\pi][/latex].
  17. False. The IVT does not work in reverse! Consider [latex](x-1)^2[/latex] over the interval [latex][-2,2][/latex].
  18. For all values of [latex]a, \, f(a)[/latex] is defined, [latex]\underset{\theta \to a}{\lim}f(\theta)[/latex] exists, and [latex]\underset{\theta \to a}{\lim}f(\theta)=f(a)[/latex]. Therefore, [latex]f(\theta)[/latex] is continuous everywhere.
  19. Nowhere

The Precise Definition of a Limit

In the following exercises (1-2), write the appropriate [latex]\varepsilon[/latex]–[latex]\delta[/latex] definition for each of the given statements.

  1. For every [latex]\varepsilon > 0[/latex], there exists a [latex]\delta > 0[/latex] so that if [latex]0 < |t-b| < \delta[/latex], then [latex]|g(t)-M| < \varepsilon[/latex]
  2. For every [latex]\varepsilon > 0[/latex], there exists a [latex]\delta > 0[/latex] so that if [latex]0 < |x-a| < \delta[/latex], then [latex]|\phi(x)-A| < \varepsilon[/latex]
  3. [latex]\delta \le 0.25[/latex]
  4. [latex]\delta \le 2[/latex]
  5. [latex]\delta \le 1[/latex]
  6. [latex]\delta < 0.3900[/latex]
  7. Let [latex]\delta =\varepsilon[/latex]. If [latex]0 < |x-3| < \varepsilon[/latex], then [latex]|x+3-6|=|x-3| < \varepsilon[/latex].
  8. Let [latex]\delta =\sqrt[4]{\varepsilon}[/latex]. If [latex]0 < |x| < \sqrt[4]{\varepsilon}[/latex], then [latex]|x^4|=x^4 < \varepsilon[/latex].
  9. Let [latex]\delta =\varepsilon^2[/latex]. If [latex]5-\varepsilon^2 < x < 5[/latex], then [latex]|\sqrt{5-x}|=\sqrt{5-x} < \varepsilon[/latex].
  10. Let [latex]\delta =\varepsilon/5[/latex]. If [latex]1-\varepsilon/5 < x < 1[/latex], then [latex]|f(x)-3|=5x-5 < \varepsilon[/latex].
  11. Let [latex]\delta =\sqrt{\frac{3}{N}}[/latex]. If [latex]0 < |x+1| < \sqrt{\frac{3}{N}}[/latex], then [latex]f(x)=\frac{3}{(x+1)^2} > N[/latex].
  12. [latex]f(x)-g(x)=f(x)+(-1)g(x)[/latex]
  13. Answers may vary.