Chapter 2: Answers 2 Jack K. Cohen Colorado School of Mines

1. f[x_] :=  x^3 + 3x^2
   f[2]
       20
       
   f[x] /. x -> 2
       20
       
   f[t]
          2    3
       3 t  + t
       
    f[cat]
            2      3
       3 cat  + cat
       
   Table[f[n], {n, 1, 5}]
       {4, 20, 54, 112, 200}
       
   Plot[f[x], {x, -2, 2}]
       See Figure 1.
   

   Figure: Plot of x³ +3x².

   
2. $\lim_{{x \rightarrow 0}}^{}$ ${\frac{{\sqrt{x + 25} - 5}}{{x}}}$ is 1/10.
   1. Figure 3 shows the output from

      Plot[(Sqrt[x + 25] - 5)/x, {x, -2, 2}, PlotRange -> {0, .5}]

      Figure: Plot for problem 2.

      

   2. Here is the output from

      Table[{x, (Sqrt[x + 25] - 5)/x}, {x, -1/5, 1/5, 1/20}] //N //TableForm

          -0.2    0.100201
          -0.15   0.10015
          -0.1    0.1001
          -0.05   0.10005
          0       Indeterminate
          0.05    0.09995
          0.1     0.0999002
          0.15    0.0998504
          0.2     0.0998008
      

      It may seem more natural to ``force'' to use decimal arithmetic during the calculation instead of evoking N afterwards, perhaps like this:

      Table[{x, (Sqrt[x + 25] - 5)/x}, {x, -.2, .2, .05}] //TableForm.

      However, with in-exact arithmetic, the user has to be alert enough to reject output that is too close to the singular point:

      -0.2             0.100201
      -0.15            0.10015
      -0.1             0.1001
      -0.05            0.10005
                 -17
      -1.38778 10      0.
      0.05             0.09995
      0.1              0.0999002
      0.15             0.0998504
      0.2              0.0998008
      

   3. If Example 12 seems obscure, ask your instructor for help.
   4. As indicated, $\sqrt{{x + 25}}$ $\approx$ 5 + x/10, so we can replace the given limit by $\lim_{{x \rightarrow 0}}^{}$ ${\frac{{(5 + x/10) -5}}{{x}}}$ = 1/10.
   5. Running the indicated code gives:

      0.1       0.0999002
      0.01      0.09999
      0.001     0.099999
      0.0001    0.0999999
      0.00001   0.1
      

      With x -> a - 1/10ˆk, we get:

      -0.1       0.1001
      -0.01      0.10001
      -0.001     0.100001
      -0.0001    0.1
      -0.00001   0.1
      

      Note: We have supplied a function called LimitTable that uses this code as its core.

   6. Your opinion is more important than mine.

3. (2.2.8) $\lim_{{t \rightarrow 2}}^{}$ ${\frac{{t^2 + 2t - 5}}{{t^3 - 2t}}}$ = 3/4.

4. (2.2.22) $\lim_{{x \rightarrow 9}}^{}$ ${\frac{{3 - \sqrt{x}}}{{9 - x}}}$ = 1/6.

5. (2.2.32) $\lim_{{x \rightarrow 1}}^{}$ ${\frac{{x^4 - 1}}{{x - 1}}}$ = 4.

6. (2.2.35) $\lim_{{x \rightarrow 0}}^{}$ ${\frac{{1}}{{x}}}$$\left(\vphantom{ \frac{1}{x + 5} - \frac{1}{5} }\right.$${\frac{{1}}{{x + 5}}}$ - ${\frac{{1}}{{5}}}$$\left.\vphantom{ \frac{1}{x + 5} - \frac{1}{5} }\right)$ = - 1/10.

7. (2.2.36) $\lim_{{x \rightarrow 8}}^{}$ ${\frac{{x^{2/3} - 4}}{{x - 8}}}$ = 1/3. To do this algebraically is a bit trickey. Factor the numerator: (x^1/3 -2)(x^1/3 + 2) and the denominator: (x^1/3 -2)(x^2/3 +2x^1/3 + 4). Then cancel the common term and substitute x = 8.