Calculus- Early Transcendentals, 2021a

194 Applications of Derivatives
This could be anything: it depends on how small and how large each piece of the function turns out to be.
As defined earlier, this is a type of ±“0 · ∞”, which is indeterminate. So we can in fact apply L’Hôpital’s
Rule after re-writing it in the form ∞ ∞ :
xlnx = lnx
1/x = lnx
x −1 .
Now as x approaches zero, both the numerator and denominator approach infinity (one −∞ and one +∞,
but only the size is important). Using L’Hôpital’s Rule:
lnx 1/x
lim = lim = lim
x→0 + x−1 x→0 + −x−2 x→0 + 1
x (−x2 )= lim −x = 0.
x→0 +
One way to interpret this is that since lim xlnx = 0, the x approaches zero much faster than the lnx
x→0 +
approaches −∞.
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Finally, we illustrate how a limit of the type “1 ∞ ” can be indeterminate.
Example 5.39: L’Hôpital’s Rule
Evaluate lim
x→1 + x1/(x−1) .
Solution. Plugging in x = 1 (from the right) gives a limit of the type “1 ∞ ”. To deal with this type of limit
we will use logarithms. Let
L = lim
x→1 x1/(x−1) .
+
Now, take the natural log of both sides:
(
lnL = lim ln x 1/(x−1)) .
x→1 +
Using log properties we have:
lnx
lnL = lim
x→1 + x − 1 .
The right side limit is now of the type 0/0, therefore, we can apply L’Hôpital’s Rule:
lnx
lnL = lim
x→1 + x − 1 = lim 1/x
x→1 + 1 = 1
Thus, lnL = 1 and hence, our original limit (denoted by L) is:L = e 1 = e. Thatis,
L = lim
x→1 + x1/(x−1) = e.
In this case, even though our limit had a type of “1 ∞ ”, it actually had a value of e.
♣
Exercises for 5.5
Compute the following limits.