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

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

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

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:

Limit[h[x], x -> 5]

2.

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

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

(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

Limit[g[x], x -> 4]

(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

Limit[g[x], x -> 0]

3

Clear[h, g]

f[x_] := x Cos[x]

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

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

Cos[0]

Limit[f[x], x -> 0]

4.

Clear[f]

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

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

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

Limit[f[t], t -> 4]

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