Karl's Calculus Tutor - Solution to Exercise 8.1-6

Solution to Exercise 8.1-6KCT logo

© 1999 by Karl Hahn

The problem was to find the limit of 

            A0 + A1x
    lim             
   x  > ¥   B0 + B1x
Where A0, A1, B0, and B1 are all constants, and A1B1 ¹ 0.

Note that the last statement -- the one about the quantity A1B1 -- implies that neither A1 nor B1 can be equal to zero. Think about it for a second.

Step 1: Confirm that the magnitudes of both numerator and denominator grow without limit. Since B0 is finite and B1 ¹ 0, we know that you can always choose x large enough so that B1x will overwhelm B0. Indeed, it is clear that the magnitude of B1x can be made arbitrarily large by choosing a large enough x. That takes care of the denominator. You have the exactly same story for the numerator, except that you replace the B's with A's. This means that this limit qualifies for the version of L'Hopital where both numerator and denominator go to infinity.

Step 2: Take the derivative of both numerator and denominator. These two are pretty easy derivatives. In the numerator you have A0, which is a constant, so its derivative is zero. And you have A1x. Its derivative is simply A1. So the derivative of the numerator is simply A1. Likewise the derivative of the denominator is simply B1.

Step 3: Apply L'Hopital's Rule. Which simply means that you replace the numerator and denominator with their respective derivatives. 

            A0 + A1x                A1
    lim               =   lim       
   x  > ¥   B0 + B1x     x  > ¥   B1

Step 4: Take the limit. This is an especially easy limit to take since x does not appear anywhere in the right-hand limit expression. As a result, you get that the limit is simply the expression itself (provided that neither A1 nor B1 are zero): 

            A0 + A1x       A1
    lim               =    
   x  > ¥   B0 + B1x      B1

See if you can figure out for yourself what happens with this limit when you retain the condition that B1 ¹ 0, but allow A1 to be zero. Observe that the magnitude of the numerator no longer goes to infinity in this case, so L'Hopital no longer applies. Can you show that the limit is still A1/B1 despite this?

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