Karl's Calculus Tutor - Solution to Exercise 8.1-9

Solution to Exercise 8.1-9KCT logo

© 1999 by Karl Hahn

The problem is to use the result of problem 8.1-8 together with induction and L'Hopital's rule to prove that 

    lim     xn e-x  =  0
   x  > ¥
for any natural number, n.

You will recall that the result of problem 8.1-8 was that you proved that 

    lim     x e-x  =  0
   x  > ¥
using L'Hopital's rule.

You will also recall that there were two possible ways in problem 8.1-8 that you could have set up L'Hopital's rule, only one of which led to a solution. In this problem there are also two possible ways you can set up L'Hopital's rule, but here both of them lead to a solution. If you used either of these methods you get full credit.

Let's begin by remembering what it means to prove something by induction. The principle is that if you can step onto the first rung of a ladder, and from any rung you can step up to the next rung, then you can get to every rung of the ladder. In problem 8.1-8 you already proved that you can get to the first rung of this ladder. That is you proved 

    lim     xn e-x  =  0
   x  > ¥
for n = 1.

Now we assume that we have made it to the nth rung of the ladder and we prove that if we can do that, we can also make it to the n+1st rung.

Here is the first method of doing that. It relies on the identity, 

            1
   e-x  =     
           ex
just as we did in problem 8.1-8. So we are trying to prove that 
            xn+1
    lim           =  0
   x  > ¥    ex
given that 
            xn
    lim         =  0
   x  > ¥   ex
Observe that both numerator and denominator of the above first limit above go to infinity as x goes to infinity. So L'Hopital's rule applies. The derivatives of numerator and denominator are (n+1)xn and ex respectively. So we have 
            xn+1              (n+1)xn
    lim           =   lim            
   x  > ¥    ex      x  > ¥     ex
But this is just n+1 times something that we have already assumed is zero. And anything times zero is zero, which completes the proof.

I promised you a different approach also. We can do this by setting up L'Hopital's rule upside down. The following is the same limit as the original problem: 

            e-x
    lim         
   x  > ¥    1
               
             xn
Observe that both numerator and denominator go to zero as x goes to infinity. So L'Hopital's rule applies. The derivatives of numerator and denominator are -e-x and -n/xn+1 respectively. So you have 
             e-x                -e-x
    lim            =   lim           
   x  > ¥     1       x  > ¥     -n
                                    
              xn                xn+1
This means that if the limit to the left of the equal goes to zero, then so must the limit to the right of the equal. But saying that the limit to the right of the equal is zero is the same as saying 
            1
    lim       xn+1 e-x  =  0
   x  > ¥   n
We know that 1/n ¹ 0. For the a product to be zero, one of its factors must be zero. Therefore it must be true that 
    lim     xn+1 e-x  =  0
   x  > ¥
Which proves that from the nth rung of this ladder you can step to the n+1st rung. And that completes this version of the proof.

Return to Main Text

email me at hahn@netsrq.com