On-line Math 21

4.4  l'Hôpital's Rule

Example 5

lim x® ¥  æ ç è 1+ 1 x ö ÷ ø x

Solution

Again, this is not in one of the forms 0/0 or ¥/¥; it has to be transformed into one of those forms.

First we deal with the natural logarithm of this expression. Then we will use the fact that

lim x® a  ef(x) = elimx® af(x),

which is true because the exponential is a continuous function, to show the limit we are really after.

lim x® ¥  ln æ ç è æ ç è 1+ 1 x ö ÷ ø x   ö ÷ ø = lim x® ¥  xln æ ç è 1+ 1 x ö ÷ ø = lim x® ¥  ln æ ç è 1+ 1 x ö ÷ ø 1 x .

The first step there just brought the exponent out of the logarithm, using the standard rules for logarithms. This changed the type of indeterminate form from 1^¥ to ¥·0 . Then, in a more standard step, the ¥·0 form is converted to 0/0 . Now, at that stage there was a choice, to put one or the other of those terms in the denominator. But it makes more sense to put the simpler term in the denominator, since you have to invert it. Inverting a complicated expression gives you a mess to differentiate. Inverting a simple expression is not as likely to cause trouble.

Continuing, applying l'Hôpital's rule and not forgetting the chain rule,

lim x® ¥  ln æ ç è æ ç è 1+ 1 x ö ÷ ø x   ö ÷ ø = lim x® ¥  ln æ ç è 1+ 1 x ö ÷ ø 1 x = lim x® ¥  æ ç è 1 1+[1/x] ö ÷ ø æ ç è 0- 1 x2 ö ÷ ø - 1 x2 = lim x® ¥  æ ç è 1 1+[1/x] ö ÷ ø ( 1) 1 = 1.

But we're not yet done. We found the limit of the logarithm. To get back to the original limit,

lim x® ¥  æ ç è 1+ 1 x ö ÷ ø x   = lim x® ¥  e( ln( 1+[1/x]) x) = e( limx® ¥ln( ( 1+[1/x]) x) ) = e( 1) = e.

Copyright (c) 2000 by David L. Johnson.