On-line Math 21

4.4  l'Hôpital's Rule

Example 4

lim x® ¥  æ è Ö   x2+x+1   - Ö   x2-x   ö ø

Solution

This is not in one of the forms 0/0 or ¥/¥; it has to be transformed into one of those forms. For this one, the trick is to conjugate; that is, multiply numerator and denominator by the sum of those square roots. This trick was used before, in an example in the section on finding the derivative by the definition.

lim x® ¥  æ è Ö   x2+x+1   - Ö   x2-x   ö ø = lim x® ¥  æ è Ö   x2+x+1   - Ö   x2-x   ö ø æ è Ö   x2+x+1   + Ö   x2-x   ö ø æ è Ö   x2+x+1   + Ö   x2-x   ö ø = lim x® ¥  (x2+x+1)-(x2-x) æ è Ö   x2+x+1   + Ö   x2-x   ö ø = lim x® ¥  2x+1 æ è Ö   x2+x+1   + Ö   x2-x   ö ø .

However, once we have done that, we no longer need l'Hôpital's rule! All we need to do here is divide out top and bottom by the largest common power of x (in this case, x itself), and bring that power inside the square roots in the denominator:

lim x® ¥  æ è Ö   x2+x+1   - Ö   x2-x   ö ø = lim x® ¥  ( 2x+1) 1 x æ è Ö   x2+x+1   + Ö   x2-x   ö ø 1 x = lim x® ¥  2+ 1 x æ è Ö   1+[1/x]+[1/(x2)]   + Ö   1-[1/x]   ö ø .

It is now standard to see that the two terms in the denominator each goes to 1, and the numerator goes to 2, leaving a final limit of 1,

lim x® ¥  æ è Ö   x2+x+1   - Ö   x2-x   ö ø = 1.

Now, you could apply l'Hôpital's rule to

lim x® ¥  2x+1 æ è Ö   x2+x+1   + Ö   x2-x   ö ø ,

but it just doesn't help:

lim x® ¥  2x+1 æ è Ö   x2+x+1   + Ö   x2-x   ö ø = lim x® ¥  2 æ ç è 2x+1 2Ö{x2+x+1} + 2x-1 2Ö{x2-x} ö ÷ ø ,

leaving you with two limits that are essentially as complicated as where we were before. This often happens with l'Hôpital's rule problems, not every problem is actually simplified by this technique.

Copyright (c) 2000 by David L. Johnson.