MATH 1010E University Mathematics
Lecture Notes (week 8)
Martin Li
1 LHospitals Rule
Another useful application of mean value theorems is LHospitals Rule. It
helps us to evaluate limits of indeterminate forms such as 00 . Lets look
at the following example. Recall that we have proved in week 3 (using the
sandwich theorem and a geometric argument)
sin x
lim = 1.
x0 x
We say that the limit above has indeterminate form 00 since both the numer-
ator and denominator goes to 0 as x 0. Roughly speaking, LHospitals
rule says that under such situation, we can differentiate the numerator and
denominator first and then take the limit. The result, if exists, should be
equal to the original limit. For example,
(sin x)0 cos x
lim 0
= lim = 1,
x0 (x) x0 1
which is equal to the limit before we differentiate!
Theorem 1.1 (LHospitals Rule) Let f, g : (a, b) R be differentiable
functions in (a, b) and fix an x0 (a, b). Assume that
(i) f (x0 ) = 0 = g(x0 ).
f 0 (x)
(ii) limxx0 g 0 (x) = L (i.e. the limit exists and is finite).
Then, we have
f (x) f 0 (x)
lim = lim 0 = L.
xx0 g(x) xx0 g (x)
Example 1.2 Consider the limit
sin2 x
lim ,
x0 1 cos x
this is a limit of indeterminate form 00 . Therefore, we can apply LHospitals
Rule to obtain
sin2 x (sin2 x)0
lim = lim ,
x0 1 cos x x0 (1 cos x)0
1
if the limit on the right hand side exists. Since the right hand side is the
same as
2 sin x cos x
lim = lim (2 cos x) = 2.
x0 sin x x0
sin2 x
Therefore, we conclude that limx0 1cos x = 2.
Exercise: Calculate the limit in Example 1.2 without using LHospitals
Rule (hint: sin2 x = 1 cos2 x).
Sometimes we have to apply LHospitals Rule a few times before we can
evaluate the limit directly. This is illustrated by the following two examples.
Example 1.3 Consider the limit
x sin x
lim ,
x0 x3
this is of the form 00 . Therefore, by LHospitals rule
x sin x 1 cos x
lim 3
= lim ,
x0 x x0 3x2
if the right hand side exists. The right hand side is still in the form 00 ,
therefore we can apply LHospitals Rule again
1 cos x sin x
lim 2
= lim ,
x0 3x x0 6x
if the right hand side exists. But now the right hand side can be evaluated:
sin x 1 sin x 1
lim = lim = .
x0 6x 6 x0 x 6
As a result, if we trace backwards, we conclude that the original limit exists
and
x sin x 1
lim 3
= .
x0 x 6
Example 1.4 Consider the limit
ex x 1
lim .
x0 1 cosh x
Applying LHospitals Rule twice, we can argue as in Example 1.3 that
ex x 1 ex 1 ex 1
lim = lim = lim = = 1.
x0 1 cosh x x0 sinh x x0 cosh x 1
2
After seeing these examples, let us now go back to give a proof of
LHospitals Rule.
Proof of LHospitals Rule: Recall Cauchys Mean Value Theorem which
says that
f (b) f (a) f 0 ()
= 0
g(b) g(a) g ()
for some (a, b). Therefore, since f (x0 ) = g(x0 ) = 0, we have
f (x) f (x) f (x0 ) f 0 ()
= = 0
g(x) g(x) g(x0 ) g ()
for some between x and x0 . Notice that as x x0 , we must also have
x0 . Therefore, we have
f (x) f 0 ()
lim = lim 0 .
xx0 g(x) x0 g ()
This proves the LHospitals Rule.
2 Other indeterminate forms
When we evaluate limits, there are other possible indeterminate forms,
for example
0
, 0 , , 00 . (2.1)
0
Note that these forms above are just formal expressions which does not have
very precise mathematical meanings as is not a real number.
Convention: We distinguish two infinities by writing
:= + and := .
Remark 2.1 Not all expressions involving 0 and would result in an in-
determinate form. For example,
0 = 0, = , + = , = .
In this section, we will see that all the indeterminate forms in (2.1)
can actually be rewritten into the standard form 00 . Symbolically we have
1/0 = . Therefore,
1 0
0=0 = ,
0 0
3
1/0 0
= = .,
1/0 0
0
00 = exp(0 ln 0) = exp(0 ()) = exp( ).
0
We should emphasize that the calculations above are just formal. They
indicate the general idea of transforming the limits rather than actual arith-
metic of numbers. Using these ideas, we can actually handle all the deter-
minate forms in (2.1) by the LHospitals Rule. We have
Theorem 2.2 (LHospitals Rule) The same conclusion holds if we re-
place (i) by
lim f (x) = = lim g(x).
xx0 xx0
Remark 2.3 The theorem also holds in the case x0 = and for one-
sided limits as well.
We postpone the proof of Theorem 2.2 until the end of this section but
we will first look at a few applications.
Example 2.4 Consider the one-side limit
lim x ln x.
x0+
This is of the form 0 (). However, we can rewrite it as
ln x
x ln x = ,
1/x
which is of the form as x 0+ . Therefore, we can apply Theorem 2.2
to conclude that
ln x 1/x
lim = lim = lim (x) = 0.
x0+ 1/x x0+ 1/x2 x0+
Therefore, we have limx0+ x ln x = 0. In words, this means that as x 0+ ,
the linear function x is going to 0 faster than the logarithm function ln x
going to .
Example 2.5 Sometimes we have to apply LHospitals Rule a few times.
For example,
x2 2x 2
lim x = lim x = lim x = 0.
x+ e x+ e x+ e
4
Similarly, we can prove that
xk
lim = 0, for any k.
x+ ex
In other words, as x +, the exponential function ex is going to faster
than any polynomial of x.
The following example shows that LHospitals Rule may not always
work:
sinh x cosh x sinh x
lim = lim = lim ,
x cosh x x sinh x x cosh x
which gets back to the original limit we want to evaluate! So LHospitals
Rule leads us nowhere in such situation. For this example, we have to do
some cancellations first,
sinh x ex ex 1 e2x
lim = lim x = lim = 1.
x cosh x x e + ex x 1 + e2x
3 Some tricky examples of LHospitals Rule
Sometimes it is not very obvious how we should transform a limit into a
standard indeterminate form.
Example 3.1 Evaluate that limit
1
lim x sin .
x x
We can choose to transform it to either
1 sin(1/x) 1 x
x sin = or x sin = .
x 1/x x 1/ sin(1/x)
The first one has the form 00 and the second one has the form
as
x . Therefore, we can apply LHospitals Rule to both cases. For the
first case, we have
sin(1/x) 12 cos 1 1
lim = lim x 1 x = lim cos = 1.
x 1/x x x2 x x
However, for the second case, we have
x 1
lim = lim ,
x 1/ sin(1/x) x 1 cos(1/x)
x2 sin2 (1/x)
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which doesnt seem to simplify after LHospitals Rule. Therefore, sometimes
we have to choose a good way to transform the limit before we apply the
LHospitals Rule. A general rule of thumb here is that the expression should
get simpler after taking the derivatives.
Example 3.2 Evaluate the limit
1 1
lim .
x0 sin x x
This limit has the indeterminate form , which we havent men-
tioned. There is in fact no general way to evaluate limits of such forms. But
for this particular example, we can transform it as
1 1 x sin x
= ,
sin x x x sin x
which has the standard indeterminate form 00 . Therefore, we can apply
LHospitals Rule a few times to get
x sin x 1 cos x sin x
lim = lim = lim = 0.
x0 x sin x x0 sin x + x cos x x0 2 cos x x sin x
Example 3.3 Evaluate the limit
1
lim x x .
x
Recall that if a > 0, b are real numbers, we define ab := exp(b ln a).
Therefore,
1 1 ln x 1/x
lim x x = lim exp ln x = exp lim = exp lim = e0 = 1.
x x x x x x 1
Note that we can move the limit into the function exp since the exponential
function exp is continuous.
We end this section with a proof of Theorem 2.2.
Proof of Theorem 2.2 : The idea is that if f (x0 ) = = g(x0 ), then we have
1 1
f (x0 ) = 0 = g(x0 ) . Therefore, we can apply LHospitals Rule to conclude
that
f (x) 1/g(x) g 0 (x)/g(x)2
lim = lim = lim .
xx0 g(x) xx0 1/f (x) xx0 f 0 (x)/f (x)2
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The right hand side is
f (x) 2 f (x) 2
0
g (x) f (x)2 g 0 (x) g 0 (x)
lim = lim 0 lim = lim 0 lim .
xx0 f 0 (x) g(x)2 xx0 f (x) xx0 g(x) xx0 f (x) xx0 g(x)
Therefore, we have
2
g 0 (x)
f (x) f (x)
lim = lim 0 lim .
xx0 g(x) xx0 f (x) xx0 g(x)
Canceling and moving terms around, we obtain
g 0 (x) 1 f 0 (x)
f (x)
lim = lim 0 = lim 0 .
xx0 g(x) xx0 f (x) xx0 g (x)
This proves the theorem.
Question: Spot the gaps in the above proof. Can you fix them?
4 Indefinite Integral
Now, we know how to differentiate a function f (x) to get a new function
f 0 (x). We want to ask whether we can reverse the process.
Question: Given a function f : (a, b) R (say differentiable), can we find
another function F : (a, b) R such that
F 0 (x) = f (x)
for all x (a, b)?
Lets try to understand the above question by some simple examples.
Example 4.1 Suppose f (x) = ex . Can we solve for F (x) such that F (x) =
f (x) = ex ? Well we know that
F (x) = ex
is a solution since (ex )0 = ex . Are there any other solutions? Yes, for
example,
F (x) = ex + 1
is another solution. In fact, for any constant C R,
F (x) = ex + C
is a solution. Are these all the solutions then? The answer is indeed YES!
7
Question: Show that if there are two solutions F1 (x) and F2 (x) such
that F10 (x) = f (x) = F20 (x) then
F1 (x) = F2 (x) + C for some constant C.
We make the following definition.
Definition 4.2 If F (x) is a differentiable function such that F 0 (x) = f (x),
we say that F (x) is a primitive function of f (x) and
Z
f (x) dx := F (x) + C
is said to be the indefinite integral of f (x). Here, C R is an arbitrary
constant called the integration constant.
For example, since we know that (ex )0 = ex and x0 = 1,
Z
ex dx = ex + C,
Z
1 dx = x + C.
Proposition 4.3 We can evaluate some of the elementary indefinite inte-
grals.
Z
1. cos x dx = sin x + C.
Z
2. sin x dx = cos x + C.
xn+1
Z
3. xn dx = + C for any real number n 6= 1.
n+1
Z
1
4. dx = ln |x| + C.
x
The following property helps us evaluate a much larger class of indefinite
integrals.
Proposition 4.4 (Linearity) Indefinite integrals are linear:
Z Z Z
1. [f (x) g(x)] dx = f (x) dx g(x) dx.
8
Z Z
2. kf (x) dx = k f (x) dx for any constant k.
Using just Proposition 4.3 and 4.4, we can evaluate a lot of indefinite
integrals. First, we know how to compute the indefinite integrals of any
polynomials. For example,
Z Z Z Z
4 4
(x 3x + 7) dx = x dx 3 x dx + 7 1 dx
5 2
x x
= + C1 3 + C2 + 7(x + C3 )
5 2
x5 3x2
= + 7x + (C1 3C2 + 7C3 )
5 2
x5 3x2
= + 7x + C
5 2
where C is ANY constant. Note that in the end, we can simply group all
the constants together to form a single constant C since all these constants
are arbitrary.
We can also evaluate the indefinite integrals of some rational functions.
For example,
(x + 2)2
Z Z 2
x + 4x + 4
dx = dx
x x
Z
4
= x+4+ dx
x
Z Z Z
1
= x dx + 4 dx + 4 dx
x
x2
= + 4x + 4 ln |x| + C.
2
The powers in the rational function do not need to be integers in some
cases.
5x2 + x + 3
Z Z
3 1
dx = (5x 2 + 1 + 3x 2 ) dx
x
x5/2 x1/2
= 5 +x+3 +C
5/2 1/2
5 1
= 2x 2 + x + 6x 2 + C.
Proposition 4.5 We can evaluate the following indefinite integrals of trigono-
metric functions.
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Z
1. sec2 x dx = tan x + C.
Z
2. csc2 x dx = cot x + C.
Z
3. sec x tan x dx = sec x + C.
Z
4. csc x cot x dx = csc x + C.
Z
5. sec x dx = ln | sec x + tan x| + C.
Z
6. csc x dx = ln | csc x + cot x| + C.
The first four integrals are easy. The last two will be proved after we
learn t-substitution in the next class.
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