March 06, 2006 7:35 am
Larson ch 1.3, p. 64 - 66
1.

In[1]:=

h[x_] := x^2 - 5x

In[2]:=

Plot[h[x], {x, 0, 10}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_3.gif]

From the graph it is clear that the limit of h as x approaches 5 is 0. Direct substitution yields the same answer. Here is the command for Mathematica:

In[4]:=

Limit[h[x], x5]

Out[4]=

0

2.

In[5]:=

g[x_] := 12 (Sqrt[x] - 3)/(x - 9)

In[9]:=

Plot[g[x], {x, 0, 6}, PlotStyleRed] ;

[Graphics:HTMLFiles/index_8.gif]

(a) From the graph it is clear that the limit of the function g as x approaches 4 is about 2.4. Direct substitution gives a value of (-12)/(-5) = 2.4. Mathematica gives

In[8]:=

Limit[g[x], x4]

Out[8]=

12/5

(b) From the graph it is clear that the limit of the function g as x approaches 0 is about 4. Direct substitution gives (-36)/(-9) = 4. Mathematica gives

In[10]:=

Limit[g[x], x0]

Out[10]=

4

3

In[11]:=

Clear[h, g]

In[12]:=

f[x_] := x Cos[x]

In[13]:=

Plot[f[x], {x, -1, 1}, PlotStyleBlue]

[Graphics:HTMLFiles/index_16.gif]

Out[13]=

⁃Graphics⁃

Looking at the graph of f, the limit of f as x approaches 0 is 0. Direct substitution gives (0)(1) = 0. Mathematica gives

In[14]:=

Cos[0]

Out[14]=

1

In[15]:=

Limit[f[x], x0]

Out[15]=

0

4.

In[16]:=

Clear[f]

In[17]:=

f[t_] := t Abs[t - 4]

In[18]:=

Plot[f[t], {t, 0, 10}, PlotStyleRed] ;

[Graphics:HTMLFiles/index_25.gif]

Wow - that is a neat graph! I was not expecting that (I'm not sure what I was expecting - I must improve my intuition).
The limit as t approaches 4 appears to be 0 but the curve is not continuous at that point. Direct substitution yields (4)(0) = 0. Mathematica gives

In[19]:=

Limit[f[t], t4]

Out[19]=

0

This was a bit of a surprise - I thought the limit might not exist, but thinking about it, it is the derivative that will not exist.
8:30 am

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