6. Limits Again!
Introduction
You probably remember that, way back in Calc 1, you defined a derivative as a
certain limit. Well, in this chapter, we’re going to find limits using derivatives.
One of the earliest textbooks in Calculus, written hardly a decade after
Newton and Leibnitz invented Calculus, was by a woman, Maria Agnesi. A
collaborator of Ms Agnesi, and possibly her teacher, was Guillaume François
Antoine, Marquis de l'Hôpital. (“Guillaume” is French for William, so today we
would call him ‘Bill’.) The method I’m going to describe is called L’Hôpital’s
Rule. (His father was a field doctor in the French Army, or something like
that.)
6.1 Limits of the form 0/0
This gives us the main theorem of this chapter:
Theorem – L’Hopital’s Rule: If f and g have any number of
derivatives near the point x = a, and if both f and g are zero at x = a,
then
( )
′( )
lim
= lim
,
→
→
( )
′( )
provided, of course, that the last limit exists, or even if it is infinity.
From Calc I, you probably remember that if f(x) has at least a second
derivative near a point x = a, that the function f, near that point, has a tangentfunction that has the form ( ) = ( ) + ( − ) ∙ ( ). As you can see,
( ) is clearly a line with slope ( ), and it is also obvious that
( ) = ( ). For example, if you let ( ) = cos , at the point
x = ½ , we have the function and its tangent as you see here:
The proof of the result goes as follows. Near the point at which the limit is
taken, x = a, the fraction ( )/ ( ) can be written
( )
( ) + ( − ) ∙ ( ) + ( )( − )
=
.
( )
( ) + ( − ) ∙ ( ) + ( )( − )
If the limit is a 0/0 – form, both f(a) and g(a) = 0, so the fraction reduces to:
( ) ( − ) ∙ ( ) + ( )( − )
=
,
( ) ( − ) ∙ ( ) + ( )( − )
and after a little cancelling, you get
( )
( )+
=
( )
( )+
which clearly approaches
1
+
2
where the unknown function
−
1
∙
2
1
+ ( )
2
−
( )
.
Note: This isn’t really a deep result; it is just a convenient way of calculating
interesting limits. Also, we don’t need f and g to have all their derivatives; we
only need them to have at least one more derivative than you use in the
calculation.
Consider the limit
−
−2
lim
.
→
− sin
As x approaches zero, the numerator and denominator are both perfectly
continuous, and both approach zero. This is an example of what is called a
“0/0 - Form” which appears to be an abbreviation of a limit of the form zero-
What is not so obvious is that, close to the point of tangency, x = ½, the
difference between the tangent function and the original function is almost
exactly a quadratic, that is, a parabola, as shown on the right, above. It isn’t
exactly a parabola, but it is very, very close. So we can write:
( )=
( )
( )( − )
,
( )( − )
1
,
2
( ) is close to constant near ½ .
1
over-zero, which would make a lot of sense. These sorts of limits are exactly
what L’Hopital’s Rule was designed to address. We would replace both the
numerator and the denominator by their (separate) derivatives.
−
−2
+
−2
lim
= lim
.
→
→
− sin
1 − cos
But wait: this limit is also a 0/0 – form! So we have to apply L’Hopital’s Rule
once again:
−
−2
+
−2
−
lim
= lim
= lim
.
→
→
→
− sin
1 − cos
sin
Oh no; we have to do it again:
+
= lim
= 2.
→
cos
6.3 Rates of Growth
It is a lot of fun to figure out how rapidly various functions go to infinity as x
approaches infinity. For instance, we know , , , √ , ln , , all
approach infinity, but is there a ranking among them?
To decide this, it is helpful to look at limits such as
lim
.
→
If the limit is zero, we could say that ex approaches infinity faster than x3. If
the limit was infinity, we would say the opposite. So, let’s find out! L’Hopital’s
Rule applies, even if the final result is infinity.
3
6
6
lim
= lim
= lim
= lim
= 0.
→
→
→
→
This means that
grows faster than . We write:
≪ .
Similarly, it is easy to show that if m, n are positive integers and m < n, then
≪ .
This is SOP for these sorts of problems; the author of a calculus text assumes
that students will be bored with just a single application of L’Hopital’s Rule
per limit, and so most limits tend to require repeated applications of it.
L’Hopital’s Rule can be extended to limits at infinity. It takes a little fancy
footwork; in fact, we have to replace the variable x by another variable, such
as x = 1/t, say, so that as t approaches 0+, x approaches . In other words, we
have to make a substitution, and use Chain Rule. But, in the end, we have the
same result.
More interestingly, the Rule can be extended to functions whose limits are
infinity. In other words, L’Hopital’s Rule applies to limits that we would say
are of the form /.
Example: Show that ln(x) grows more slowly than √ .
1
√
√
2√
lim
= lim
= lim
= lim
= ∞,
1
→ ln( )
→
→ 2√
→
2
which tells us that ln ≪ √ .
It is quite a challenge to establish something like
≪ , but it is true. A
look at the graphs of the two functions establishes it easily, but to do it with
limits is harder.
6.2 Limits of the form /
There isn’t much benefit in restating the Theorem; it is exactly the same. We
just have two functions f, g, both of which approach infinity as x approaches a.
But if their quotient approaches a limit, it will be the same as the quotient of
their derivatives.
Consider
− +2
lim
.
→
x +
This is an /-form, so L’Hopital’s Rule applies, and we get
2 −1
2
lim
= lim = 1.
→ 2 +1
→ 2
2
The gold curve is xx, the green curve
is ex.
The green curve starts off higher
than the gold curve, but at x = e,
predictably, it cuts across, and
begins to grow faster.
ln 1 +
= lim


We cancel the factor
which means, L = e.
→
lim (sec )
→
Let = lim (sec )
Take natural logarithms of both sides:
ln
= lim [cot
→
∙ ln(sec )] = lim
→
ln(sec )
.
tan
We now have a 0/0 –form, so we apply L’Hopital’s Rule:
sec ∙ tan
1
sec
ln = lim
= lim
→
→ 2 sec
2 tan ∙ sec
So
.
Since ln x has a derivative everywhere, it is certainly continuous. So at any
point in its domain, we can use the result
ln(lim → ( )) = lim → ln ( ) , provided f(x) has a limit. So, in this
case,
1
1
1
ln = ln lim 1 +
= lim ln 1 +
= lim ∙ ln 1 +
→
.
→
Now we take logarithms of both sides:
→
.
This is a limit of the form 1. We begin in the same way as in the previous
example, and name the limit L.
→
→
⎝
⎠
, and take the limit, and obtain 1. So Ln(L) = 1,
Example: Find
The quantity inside the parentheses approaches 1, while the power
approaches , hence the name “1 -form.” While 1 raised to any power is 1,
any base greater than 1 raised to the power  would be infinite. You have to
admit that the quantity inside the parentheses is actually never 1. So which
wins: the base or the exponent?
This is a classic example of an intermediate result: it is a draw. The final limit
is actually e. The limit, as you know, is one of the possible definitions of the
constant e. But we can use L’Hopital’s Rule to verify the result, more of a test
that L’Hopital’s Rule works, than of the definition.
To address this sort of problem, we first name the limit L, or anything we
please, really:
1
= lim 1 +
.
1
.
We have successfully converted the 1 -form into a 0/0-form, and we can now
apply L’Hopital’s Rule:
1
−1
1∙
⎛1 +
⎞
Ln = lim ⎜
⎟.
−1
→
6.4 Limits of the form 1 , and 00
Consider the following limit, which we have seen already:
1
lim 1 +
.
ln = ln lim 1 +
1
→
1
→
3
=√ .
1
= .
2
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4
3.
Exercises 6
lim
→
lim
→
√43 + − √55 −
√ + 3 − √15 −
tan − sin
−1−
2. lim (csc )
→ ⁄
5.
√4 +
→
Find the following limits:
1.
lim
−
1
2
4.
lim
→
∙ sin
7
6.
lim
→
5
sin −
+