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

Solution to Exercise 8.1-4KCT logo

© 1998 by Karl Hahn

The problem was to find the limit of 

    lim     x sin(1/x)
   x  > ¥
Step 1: Express this as a quotient. So you have 
                                    sin(1/x)
    lim     x sin(1/x)  =   lim             
   x  > ¥                  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 

                                          -1
                                 cos(1/x)    
            sin(1/x)                       x2
    lim               =   lim                
   x  > ¥       1        x  > ¥       -1
                                         
                x                      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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