Examples from Section 6.6
L'Hôpital's Rule

p. 496

Applying L'Hôpital's Rule

6.

(lim
            _(x -> ∞))((x  -  8  (x^2))/(12  (x^2) +  
              5  x))  =  
        (lim_(x -> ∞))((1  -  16  x)/(24  x  +  5
                ))  =  (lim_(x -> ∞))((1/x  -  16)/(24  +  5/x))  =  
        ((-16)/24) =  -(( 2)/3)

Limit[((x  -  8  (x^2))/(12  (x^2) +  5  x)), x -> ∞]

17.

\(lim
            \+\(x -> 0\)\(t \((1 - cos \((t)\))\)\)\/\(t\  - \ 
              sin \((t)\)\)\  = \ 
        \(lim\_\(x -> 0\)\ 
            \(1 - cos \((t)\) + t\ sin \((t)\)\)\/\(1\  - \ cos \((t)\)\)\  = 
          \ \(lim\_\(x -> 0\)\(2  sin \((t)\)\  + \ 
                  t\ cos \((t)\)\)\/\(sin \((t)\)\)
 = \ 
            \(lim\_\(x -> 0\)\ 
                \(3  cos \((t)\)\  - \ t\ sin \((t)\)\)\/\(cos \((t)\)\)\  = 
              \ 3\)\)\)\)

Limit[ ((t (1  -  Cos[t]))/(t  -  Sin[t])),  t -> 0]

23.

(lim_(x -> 0)) ((x (2^x))/((2^x) -  1))  =  
      (lim_(x -> 0))(((2^x) +  x (2^x) ln (2))/((2^x) 
              ln (2))) =  
        ((2^0)/((2^0) ln (2))) =  (1/(ln (2)))

Limit[((x (2^x))/((2^x) -  1)), x -> 0]

41.

(lim_(t -> ∞))(((e^t) +  (t^2))/((e^t) -  t)) =  
      (lim_(t -> ∞))(((e^t)  +  2  t)/((e^t) -  1)) =  
        (lim_(t -> ∞))(((e^t)  +  2)/(e^t)) = 
          (lim_(t -> ∞))((e^t)/(e^t)) =  1

Limit[(((E^t) +  (t^2))/((E^t) -  t)), t -> ∞]

Dividing top and bottom by (E^t)converts the expression into a form Mathematica can handle.

Limit[((1 +  (E^(-t)) (t^2))/(1 -  (E^(-t)) t)), 
      t -> ∞]

Limits involving Bases and Exponents

50.

(lim_(x -> 0))((((e^x) +  x))^(1/x))  =  
        (lim_(x -> 0))(e^(ln ((e^x) +  x)/x)) =  
          exp ((lim_(x -> 0))ln ((e^x) +  x)/x )  =  
            (e^2) since
(lim_(x -> 0))((ln ((e^x) +  x))/x)  =  
        (lim_(x -> 0)) 
            (((1/((e^x) +  x)) ((e^x) +  1) )/1) =  
          (1/((e^0) +  0)) ((e^0) +  1) = 2

Limit[((((E^x) +  x))^(1/x)), x -> 0]

Theory and Applications

61.

Find the value of c that makes the following function continuous at x  =  0.
f (x) = {({{((9  x  -  3 sin (3  x))/(5  (x^3))), , x ≠ 0}, {c, , x  =  0}})
Solution. The appropriate value for c is simply the limit, if it exists,
\(\(c\  = \ 
        \(lim\_\(x -> 0\ 
              \)\(9  x\  - \ 3\ sin \((3  x)\)\)\/\(5  x\^3\) = 
          \(lim\_\(x -> 0\ 
                \)\(9\  - \ \ 9\ cos \((3  x)\)\)\/\(15  x\^2\)\  = \ 
            \(lim\_\(x -> 0\ \)\(27\ sin \((3  x)\)\)\/\(30  x\)

              			 = \ 
              \(lim\_\(x -> 0\ \)\(81\ cos \((3  x)\)\)\/30\  = \ 
                \(81\/30 = \ \(27\/10 = \ 2.7\)\)\)\)\)\)\ \)\)

Limit[((9  x  -  3 Sin[3  x])/(5  (x^3))), x -> 0]

Grapher Explorations

72.

a)

Plot[Sin[x]^Tan[x], {x, -7, 7}]

The gaps appear where sin (x) is negative, and hence sin (((x))^(tan (x)))is not defined.

b)

Plot[Sin[x]^Tan[x], {x, 0, π}]

Note that sin (((x))^(tan (x)))is not defined at x  =  π/2, but
\(lim\_\(x -> π/2\)sin \((x)\)\^\(tan \((x)\)\)\  = \ 
        \(lim\_\(x -> π/2\)e\^\(tan \((x)\) ln \((sin \((x)\))\)\)

          			 = \ 
          \(exp \((
                lim\_\(x -> π/2\)\ 
                  \(sin \((x)\) ln \((sin \((x)\))\)\)\/\(cos \((x)\)\)\ )
                \)\  = \ \(e\^0 = \ 1\)\)\)\)
since
(lim_(x -> π/2)) 
          ((sin (x) ln (sin (x)))/(cos (x)))  =  
        (lim_(x -> π/2)) 
            ((cos (x) ln (sin (x))  +   
                cos (x))/(-sin (x)))  =  0

Limit[Sin[x]^Tan[x], x -> (π/2)]

c) Find the extreme values of sin (((x))^(tan (x)))on the interval [0, π].

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

f'[x]  // Simplify

Solve[f'[x]  ==  0]

NSolve[f'[x]  ==  0]

The usual methods are not working. We next try FindRoot[], which will use numerical methods to find the solution to f' (x)  =  0 nearest to the starting value we give it. From the graph the minimum appears to be near x  =  0.5 and the maximum near x  =  2.5.

minx = FindRoot[f'[x] == 0, {x, 0.5}]

f[x]  /.  minx

maxx = FindRoot[f'[x] == 0, {x, 2.5}]

f[x]  /.  maxx