Karl's Calculus Tutor - Solution to Exercise 8.1-4

Solution to Exercise 8.1-4


© 1998 by Karl Hahn
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The problem was to find the limit of

    lim     x sin(1/x)
   x → ∞
Step 1: Express this as a quotient. So you have


    lim     x sin(1/x)  =
   x → ∞



   lim
  x → ∞


  sin(1/x)
          
      1
       
      x
You will agree that those two limits are identical, won't you?

Step 2: Confirm that both numerator and denominator go to zero. As x gets very large,  1/x  gets very close to zero, and indeed goes to zero in the limit. So the denominator goes to zero. And since we are taking  sin(1/x)  and we have determined that  1/x  goes to zero, and since sine is a continuous function, it must be the case that  sin(1/x)  goes to zero as well as x grows without limit (i.e. as x goes to infinity). So this limit qualifies for L'Hopital's Rule.

Step 3: Take the derivatives of the numerator and denominator. The denominator is easy. Its derivative is  -1/x2.  The function,  sin(1/x),  is a composite, so we have to use the chain rule to find its derivative. When you do, you find that its derivative is

   cos(1/x) (-1/x2)

Step 4: Apply L'Hopital's Rule. That means




    lim
   x → ∞




     sin(1/x)
               =
         1
          
         x



   lim
  x → ∞


             -1
    cos(1/x)   
             x2
                
          -1
            
          x2

Step 5: Take the limit. The  -1/x2  terms in the numerator and denominator cancel. You are left with  cos(1/x).  We know that as x grows without limit,  1/x  goes to zero. Since cosine is a continuous function, the limit as x grows without limit of  cos(1/x)  must be  cos(0) = 1,  which is the answer.


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