p. 498

71.

a)

f[x_]  :=  Sin[x]^x

Plot[f[x],  {x, 0, π}]

Limit[Sin[x]^x,  x -> 0]

To get a continuous function, sin*(((x))^x)should be defined to be 1 at x  =  0.

b) Use L'Hopital's Rule

(lim_(x -> 0))sin (((x))^x)  =  
        (lim_(x -> 0))(e^(x ln (sin (x))))  =  
          exp ((lim_(x -> 0))x ln (sin (x)))  =  
            (e^0) =  1

Since,  \(lim\_\(x -> 0\)x\ ln \((sin \((x)\))\)\  = 
        \(lim\_\(x -> 0\)\(ln \((sin \((x)\))\)\ \)\/\(1/x\) = \ 
          \(lim\_\(x -> 0\)\(\(1\/\(sin \((x)\)\)\) cos 
                  \((x)\)\)\/\(\(-1\)/x^2\)\  = \ 
            \(lim\_\(x -> 0\)\(\(x\^2\) cos \((x)\)\)\/\(sin \((x)\)\)\ 
              
				 = \ \ 
              \(lim\_\(x -> 0\)\(2  x\ cos \((x)\)\  - \ \ 
                      \(x\^2\) sin \((x)\)\)\/\(cos \((x)\)\)\  = \ 0
                \)\)\)\)\)

c) The maximum of f (x)appears to be 1 at the points x  =  0 and x  =  1.7.

d) Plot f' (x).

Plot[{f[x], f'[x]},  {x, 0, π}]

The graph of f' (x) looks like it crosses the x-axis at x  =  1.6.

e) Solve f' (x)  =  0.

FindRoot[f'[x]  ==  0,  {x, 1.6}]

f[1.7]

f[1.6]

f[1.57079686120186341]

It looks like the maximum of f (x) is 1 at x  =  1.57079686120186341`. In fact, after a little thought, it is clear that sin(π/2)  =  1 so \(TraditionalForm
      \`f(π/2)\  = \ 
        \(\(sin(π/2)\)\^\(π/2\) = \ \(1\^\(π/2\)\  = \ 1\)\)\) and since  
      0  ≤  f(x)  ≤  1 we see that f(x) does indeed have a maximum of 1 at x  =  π/2  =  1.5708 .

A more interesting question is to find where f(x) has a minimum. Inspecting the above graphs leads to a guess that the minimu is near x  =  0.4.

FindRoot[f'[x]  ==  0,  {x,  0.4}]

This is near, but not equal to, π/8 = 0.392699. There is no easy way to find an x other than  x =  π/2 where the derivative is 0.

f'[x]  // Simplify

73

a)

f[k_, x_]  :=  (((x^k) -  1)/k)

p  =  
      Plot[ Log[x] ,  {x, 0, 50},  PlotStyle -> RGBColor[1, 0, 0] ]

q  =  
      Plot[{ f[-1, x],  f[- .5, x],  f[- .1, x], f[ .1, x],  
          f[ .5, x],  f[1, x]},  {x, 0, 50}]

Show[p,  q]

b)

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

Let us show (lim_(x -> 0))(((a^x) -  1)/x) =  ln (a)using L'Hopital's Rule:

(lim_(x -> 0))(((a^x) -  1)/x) =  
        (lim_(x -> 0))(((a^x) ln (a))/1) =  
          (a^0) ln (a)  =  ln (a).