Calculus 1, methods of computing limits prior to L'Hopital's Rule. Includes the sandwich theorem.

1. Computing Limits
Tyler Murphy
September 10, 2014
lim
x!a
f(x) =?
There are a variety of ways to compute limits. The method you choose is based upon
what the limit looks like. Here we will examine a few of the common ones and some short
cuts that you will get to use later in class but for now you can use to check your work.
1 If a is a number (i.e. not in

2. nity)
1.1 Compute f(a)
First, try to plug a into f(x). If f(a) is a number, then you are done. This number is L.
1.1.1 example
lim
x!2
7x 4
x 3
f(2) =
7(2) 4
2 3
=
14 4
1
= 10
1.2 When f(a) is unde

3. ned
If you get f(a) =
0
0
do some algebra to simplify f(x)
1.2.1 example
lim
x!4
x2 7x + 12
x 4
Note that f(4) = 0
0 so we can't determine the limit simply by computing f(a).
Now we factor the function and see if we can eliminate the hole we have at 4.
1

4. x2 7x + 12
x 4
=
(x 3)(x 4)
x 4
= x 3
So, after eliminating the hole we have
lim
x!4
(x 3)
Now we can plug in the value for a. 4-3 = 1. So our limit is 1.
Simply repeat this process of factoring until you don't have an unde

5. ned value when
we compute f(a).
NOTE: If you have simpli

6. ed f(x) as much as possible and f(a) is still unde

7. ned, then
the limit likely does not exist. To be certain of this, graph the function. If it is apparent
that limx!a6= limx!a+ then the limit does not exist. If, however, limx!a appears to
equal limx!a+, then we need to try something dierent.
1.3 The Sandwich Theorem
The sandwich theorem is awesome. But it can be a little confusing to get your head around.
Brie
y I'll state what the sandwich theorem states.
The sandwich theorem says that if you have 3 functions, f(x); g(x); h(x) such that
f(x) g(x) h(x) for all x, then:
lim
x!a
f(x) lim
x!a
g(x) lim
x!a
h(x)
The sandwich theorem is useful if you have a function that is a little confusing and
you're unsure how to compute the limit. If you can generate a function that is always less
than it and one that is always greater, then when those two functions meet at a, then you'll
have something like:
lim
x!a
f(x) lim
x!a
g(x) lim
x!a
h(x)
For example, if lim x ! 0f(x) = 1 and limx!0 h(x) = 1 and you know that over a
speci

8. c domain f(x) g(x) h(x), then you know that 1 limx!0 g(x) 1. So
limx!0 g(x) = 1.
Let's see how this works.
1.3.1 example
We'll use the sandwich theorem to show that:
lim
!0
sin
= 1
Using Figure 1, consider a few things we have drawn on the unit circle.
2

9. Figure 1: The Unit Circle
First, we have drawn a line from the center through the edge of the circle. This
extended radius forms an angle x with the horizontal axis. A few lines intersect that
extended radius. First, the red line, which is perpendicular to the horizontal axis extends
from the edge of the circle up to the extended radius. This forms a triangle we will refer
to as RED. Next is the triangle formed by the blue line that intersects the outer edge of
the circle at the point on the axis and the point at the extended radius. We will call this
triangle BLUE. The last thing we will consider is the slice of the pie, if you will, formed
by the extended radius and the x-axis and the arc of the circle between them. We will refer
to this as the SLICE.
Consider now how the areas of these three

10. gures relates. The Area of BLUE is the
smallest. No matter what x is, this triangle will always be smaller than SLICE because
SLICE includes BLUE and the sliver of the arc between the blue line and the circle.
Similiarly, RED will always have a larger area than SLICE because it includes SLICE
and some area outside the circle.
So we can represent these areas with an inequality.
BLUE SLICE RED
Now let's

11. nd some numeric values for each area.
The area of a triangle is 1
2 b h. So we look at the height of BLUE, drawn in black.
This line is also the side of a triangle whose hypotenuse is the radius, which is 1, since this
is the unit circle. Since sin x =
opposite
hypotenuse
, we have that sin x =
opposite
1
= opposite. So
the length of this side is sin x, which is the height of BLUE. This also makes sense with
what we know of the coordinate values on the unit circle. The point that the height of
3

12. BLUE intersects the circle has coordinates (cos x; sin x) where sin x is the y-value (height).
Note that the base of BLUE is the radius of the circle, 1.
So the area of BLUE is 1
2 1 sin x =
sin x
2
.
SLICE To

13. nd the area of the slice, we must consider how much area it takes up out
of the whole. Realize that SLICE sweeps out an area of x radians out of the 2 possible.
If x were
2 , then it would be
2
2
= 1
4 of the whole pie. So the area that x swept out would
be 1
4
th
of the total area.
In general, SLICE takes up x
2 percent of the total area. So what is the total area?
We know that the area of a circle is r2. Since r = 1, the total area is .
So the area of SLICE = x
2 = x
2 .
Now we consider the area of RED. We know that the adjacent side is length 1. We
want the opposite side. So we need to look at the tan x =
opposite
adjacent
=
opposite
1
= opposite:
So the height of RED is tan x. The base is 1. So the area of RED = 1
2 1 tan x =
tan x
2
.
We can now put these areas into our inequality.
BLUE SLICE RED
sin x
2
x
2
tan x
2
Now that we have the inequality, we need to manipulate the center to look like
sin x
x
.
(1)
sin x
2
x
2
tan x
2
(2) sin x x tan x ( Multiply by 2 to clear denominators)
(3) sin x
sin x
x
sin x
tan x
sin x
(We need a fraction with x on one side and sin x on the other)
(4) 1
x
sin x
1
cosx
(Simplify. Remember that tan x = sin x
cos x )
(5) 1
sin x
x
cos x (when you take the inverse of an inequality, you must change the signs)
To motivate step (5) consider what happens when you have 2 1
3 and you take the
reciprocals. What do you have to do to the sign to keep the statement true.
Now we have an inequality that has the function we wanted in it in the center. Now
we can take the limit of each piece. So,
lim
x!0
1 lim
x!0
sin x
x
lim
x!0
cos x
4

14. Now we evaluate those limits that we easily can.
Clearly, limx!0 1 = 1 and by inputting 0 for x we see that limx!0 cos x = 1.
So our inequality now says that
1 lim
x!0
sin x
x
1
So limx!0
sin x
x
must be 1.
That's how the Sandwich Theorem works.
2 Some Handy Identities
lim
!0
sin
= 1
lim
!0
1 cos
= 0
algebra sandwich theorem L'Hopital's rule
5