Karl's Calculus Tutor - Solution to Exercise 8.1-1

Solution to Exercise 8.1-1KCT logo

© 1998 by Karl Hahn

a) The problem was to use L'Hopital's Rule to determine the limit

           ex - 1
    lim          
   x  > 0     x

Step 1: Determine that both numerator and denominator go to zero in the limit. Since e0 = 1, it is clear that the numerator goes to zero when x approaches zero. And it is even clearer that the denominator goes to zero under those same conditions. Since both numerator and denominator are continuous and differentiable at x = 0, this quotient qualifies for L'Hopital's Rule.

Step 2: Find the derivatives of the numerator and denominator. The derivative of ex - 1 is ex. And the derivative of x is simply 1.

Step 3: Apply L'Hopital's Rule. This means that

           ex - 1             ex
    lim            =   lim      
   x  > 0     x      x  > 0   1

Step 4: Take the limit. When x is zero, the right-hand expression above clearly goes to e0 = 1. And so the limit is 1.


b) The problem was to find the limit

           ex - 1 - x
    lim              
   x  > 0      x2

Step 1: Does it qualify for L'Hopital? The numerator of this one differs from the numerator of the first one only by x. And as x approaches zero, that's no difference at all. So if ex - 1 goes to zero as x approaches zero, then so does ex - 1 - x. And there can be no doubt that x2 goes to zero as x approaches zero. Both numerator and denominator are continuous and differentiable around x = 0. So it does qualify.

Step 2: Take the derivatives of numerator and denominator. The derivative of ex - 1 - x is ex - 1. The derivative of x2 is 2x.

Step 3: Apply L'Hopital's Rule. That means that

           ex - 1 - x             ex - 1
    lim                =   lim          
   x  > 0      x2         x  > 0    2x

Step 4: Take the limit. We ended up with a numerator and denominator that are again both zero in the limit. So we need to apply L'Hopital again. But observe that the expression to the right of the equal sign above is exactly one half of the expression you were asked to take the limit of in the first problem. And you know that that limit came out to be 1. So this limit, when you have applied L'Hopital for the second time, must come out to be 1/2.


c) The problem was to find the limit

           ex - 1 - x - (1/2)x2
    lim                        
   x  > 0           x3
The steps are the same as in the previous two problems. Observe that when you take the derivatives of the numerator and denominator and then apply L'Hopital's Rule, you get
           ex - 1 - x - (1/2)x2             ex - 1 - x
    lim                          =   lim              
   x  > 0           x3              x  > 0      3x2
Observe that the expression to the right of the equal is exactly one third of the expression you were asked to take the limit of in the last example. The last example had a limit of 1/2 so this one must have a limit of 1/6.


d) The problem was to find the limit

           ex - 1 - x - (1/2)x2 - (1/6)x3
    lim                                  
   x  > 0               x4
Do you have any doubt that when you take the derivatives of numerator and denominator and then apply L'Hopital's Rule that you will get anything other than one fourth of the expression you were asked to find the limit of in the last example? The last example had a limit of 1/6, so this one must have a limit of 1/24.


So what is the pattern here? Clearly I am using the limit I get in each example to be the coefficient of the new term that I add to the numerator to get the next example. And in the denominator I am simply increasing the power of x by one with each successive example. So if there were an example e), it would be

          ex - 1 - x - (1/2)x2 - (1/6)x3 - (1/24)x4
   lim                                             
  x  > 0                      x5
And its limit would have to be one fifth of the limit we arrived at for the last example (the one with the fourth power in the denominator). Its limit was 1/24, so the limit here would be 1/120.

Recall that the limit in the first example was 1, which is the same as 1/1!. The limit in the second example was 1/(1 * 2) or 1/2!. The limit in the third example was 1/(1 * 2 * 3), or 1/3!. The limit in the fourth example was 1/4!. And if there'd been a fifth example, its limit would have been 1/5!. And yes the pattern would continue. Notice also that the pattern of reciprocal factorials continues in the numerator coefficients as well.

Now think about what would have happened with that fifth example if you had started on it without ever seeing the other four. You would have had to apply L'Hopital's rule five times. So you would have had to take the derivative of x5 five times. And at the end of that you would have ended up with a constant of 5!. Here is an interesting and useful fact you can prove for yourself using induction: If you start with xn, where n is a whole number, and you take its derivative exactly n times, you will end up with a constant of n!.

See how that fact also plays into the choice of coefficients for the numerator terms. In the nth example in this series of examples, you will have to apply L'Hopital n times. On the n-1st time, you will have to end up with a numerator of ex - 1. Why? Because the numerator must have a limit of zero on each iteration of L'Hopital. That means that the highest degree term in the numerator must end up having an n-1st derivative of 1. And since it starts out as a coefficient times xn-1, you will need that coefficient to be 1/(n-1)! in order to cancel the (n-1)! factor that you will get by taking the derivative n-1 times of xn-1. So in general the nth example must be

           ex - 1 - x - (1/2)x2 - (1/6)x3 - ... - (1/(n-1)!)xn-1
    lim                                                        
   x  > 0                           xn
and its limit would be 1/n!, which would result from applying L'Hopital n times.

We will be coming back to this line of thought in the next section. So go ahead and play with these examples until you the pattern here begins to seem intriguing.


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