MATH 1502 - FALL 2003

EXAM 1 - SOLUTIONS

1.    (20 points; 6,6,8)  Evaluate each of the following limits using L'Hopital's Rule.  In each case specify the indeterminate form that exists.

(a)    Limit(sin(3*x)/(x-Pi),x = Pi)  .    This is the indeterminant form 0/0  .   Using L'Hopital's Rule:

         Limit(sin(3*x)/(x-Pi),x = Pi)   =   Limit(3*cos(3*x)/1,x = Pi)   =   -3

(b)    Limit(ln(x)^2/x,x = infinity)  .     This is the indeterminant form infinity/infinity  .    Using L'Hopital's Rule twice

         Limit(ln(x)^2/x,x = infinity)    =   Limit([2*ln(x)*[1/x]]/1,x = infinity)    =   Limit(2*ln(x)/x,x = infinity)   =   Limit([2/x]/1,x = infinity)    =   0

(c)    Limit((1+a/x)^x,x = infinity) .     This is the indeterminant form 1^infinity  .   Using L'Hopital's Rule, taking the natural logarithm:

        Limit(x*ln(1+a/x),x = infinity)   =   Limit(ln(1+a/x)/(1/x),x = infinity)     (indeterminant form 0/0  )

              =   Limit([1/(1+a/x)]*(-a/(x^2))/(-1/(x^2)),x = infinity)         =    Limit(a*[1/(1+a/x)],x = infinity)    =    a

     ====>    Limit((1+a/x)^x,x = infinity) = exp(a)

2.   (10 points)  Find the horizontal asymptote  of the function f(x) = x*ln((x-1)/(x+1))   where x > 1.

Solution:  To find the HA, compute Limit(x*ln((x-1)/(x+1)),x = infinity) .   This is the indeterminant form infinity  * 0 .   Using L'Hopital's Rule:

    Limit(x*ln((x-1)/(x+1)),x = infinity)   =   Limit([ln((x-1)/(x+1))]/(1/x),x = infinity)    

                                    =   Limit([2/(x^2-1)]/(-1/(x^2)),x = infinity)    =   Limit(-2*x^2/(x^2-1),x = infinity)   =   -2

Thus, the line y = -2  is a horizontal asymptote

3.    (30 points; 10,10,10)   Each of the following series converges .  Find the sum.

(a)    Sum(1/(2*k*(k+1)),k = 3 .. infinity)  .   (Hint: telescoping series)

Solution:  This is a telescoping series with f(k) = 1/k  ; thus 1/(k*(k+1)) = 1/k-1/(k+1)   ===>

          Sum(1/(2*k*(k+1)),k = 3 .. infinity)   =   [1/2]*Sum(1/k-1/(k+1),k = 3 .. infinity)  

      The partial sums are:   S[n] = [1/2]*Sum(1/k-1/(k+1),k = 3 .. n)   =   [1/2]*[1/3-1/(n+1)]    -----> [1/2]*[1/3] = 1/6

(b)    Sum(5/(3^k)-7/(4^k),k = 0 .. infinity)

Solution:  This is the difference of two geometric series:

    Sum(5/(3^k)-7/(4^k),k = 0 .. infinity)   =    5*Sum((1/3)^k,k = 0 .. infinity)-7*Sum((1/4)^k,k = 0 .. infinity)   

      =    5*[1/(1-1/3)]-7*[1/(1-1/4)]    =   5*[3/2]-7*[4/3]    =   -11/6

(c)     Sum(21/(10^(2*k)),k = 1 .. infinity)

Solution:      Sum(21/(10^(2*k)),k = 1 .. infinity)   =    21*Sum(1/(10^(2*k)),k = 1 .. infinity)    =    21*Sum((1/100)^k,k = 1 .. infinity)

                                             =    [21]*(1/(1-1/100)-1)    =   21/99 = 7/33

4.   (15 points)  Use the Integral Test  to determine if the following series converges or diverges:      Sum(k/exp(k^2),k = 1 .. infinity)   

Solution:   Compare the series to the improper integral Int(x/exp(x^2),x = 1 .. infinity)   .  To evaluate this integral use IBS on the integral:  ( u = x^2   &   du = 2*x*dx )

   Int(x/exp(x^2),x = 1 .. b)   =   [1/2]*Int(exp(-u),u = 1 .. b^2)   =   [1/2]*(exp(-1)-exp(-b^2))    ---->   [1/2]*exp(-1)     as  b ----> infinity

   Since the improper integral converges, the series converges by the Integral Test.

5.   (10 points)   Use the Basic Comparison Test to show the the series Sum(1/k!,k = 1 .. infinity)    converges.   (Hint:  Show that 1/k! < (1/2)^(k-1) )

Solution:    k!  = 1*2*3* ... k > 2*2*2*2 ...  *2  (k -1 times)  ===>   1/k! < (1/2)^(k-1) .   Now the series Sum((1/2)^(k-1),k = 1 .. infinity)    is a geometric series that converges.  By the BCT,    Sum(1/k!,k = 1 .. infinity)   also converges.

6.  (15 points)  (a)  Use L'Hopital's Rule to evaluate Limit(ln(k)/k,k = infinity)   

(b)  Use (a) and the Limit Comparison Test  to determine if Sum(1/(2*k+ln(k)),k = 1 .. infinity)   converges or diverges.

Solution:  (a)  By L'Hopital's Rule,   Limit(ln(x)/x,x = infinity)   =   Limit([1/x]/1,x = infinity)    =  0;  therefore, limit(ln(k)/k,k = infinity) = 0

(b)  Use a[k] = 1/k   and b[k] = 1/(2*k+ln(k))

      Limit(a[k]/b[k],k = infinity)    =    Limit([1/k]/(1/(2*k+ln(k))),k = infinity)    

                        

                         =   Limit((2*k+ln(k))/k,k = infinity)   =   Limit(2+ln(k)/k,k = infinity)   = 2  (by (a) limit(ln(k)/k,k = infinity) = 0 )

By the LCT,  since Sum(1/k,k = 1 .. infinity)   diverges,   Sum(1/(2*k+ln(k)),k = 1 .. infinity)   diverges.