Karl's Calculus Tutor - Solution to Exercise 8.1-7

Solution to Exercise 8.1-7KCT logo

© 1999 by Karl Hahn

The problem is to use the result from problem 6 together with induction to show that if 

   P(x)  =  A0 + A1x + A2x2 + ... + Anxn
and 
   Q(x)  =  B0 + B1x + B2x2 + ... + Bnxn
and Bn ¹ 0, then 
           P(x)     An
   lim           =    
  x  > ¥   Q(x)     Bn

Recall that to prove something by induction, you have to be able to go up the ladder. That means getting onto the first rung, and from each rung getting onto the next rung.

In problem 6 you already demonstrated that you can get onto the first rung. That is you proved it for a first degree polynomial (that is, you proved it for n = 1). All that's left to do is to show that from any rung you can get to the next rung. To do this, you start by assuming that you can get to the nth rung and prove that from there you can get to the n+1st rung.

Assuming that you can get to the nth rung means assuming that the theorem is true for polynomials of degree n. Now suppose the degree were n+1. 

   P(x)  =  A0 + A1x + A2x2 + ... + Anxn + An+1xn+1
and 
   Q(x)  =  B0 + B1x + B2x2 + ... + Bnxn + Bn+1xn+1
So you seek to find 
           A0 + A1x + A2x2 + ... + Anxn + An+1xn+1
   lim                                          
  x  > ¥   B0 + B1x + B2x2 + ... + Bnxn + Bn+1xn+1
where Bn+1 ¹ 0.

Both numerator and denominator have magnitutes that go to infinity as x goes to infinity, so L'Hopital's Rule says that you can take the derivatives of both the numerator and denominator, and their quotient should have the same limit as the quotient we're after. So let's do that: 

           A1 + 2A2x + ... + nAnxn-1 + (n+1)An+1xn
   lim                                          
  x  > ¥   B1 + 2B2x + ... + nBnxn-1 + (n+1)Bn+1xn
Observe that the above is only the quotient of nth degree polynomials. And we have assumed that the theorem is true for them. So, by that assumption, the limit of the above must be 
   (n+1)An+1      An+1
              =      
   (n+1)Bn+1      Bn+1
which is what we were seeking to get when in the n+1 case. Starting with the assumption that the theorem works for polynomials of degree n we have shown that the theorem must also work for polynomials of degree n+1. And since we previously showed that it always works for first degree polynomials, induction says it must work with any degree polynomial.

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