Transcript Finding Limits Algebraically Chapter 2: Limits and Continuity What you’ll learn about • • • • • Finding a limit algebraically (or analytically) Using properties of limits Finding a.

Finding Limits Algebraically
Chapter 2: Limits and Continuity
What you’ll learn about
•
•
•
•
•
Finding a limit algebraically (or analytically)
Using properties of limits
Finding a limit using the sandwich (or squeeze) theorem
Two special limits
Limits of piecewise functions
Direct Substitution
In the last lesson, we learned that the limit of f(x) as x approaches c
doesn’t depend on the actual value of f at x = c. It often happens,
however, that the limit is actually f(c). In these cases, the limit can be
evaluated by direct substitution. In other words,
lim 𝑓 𝑥 = 𝑓(𝑐) (just substitute c for x)
𝑥→𝑐
For example, lim 3𝑥 = 3 · 2 = 6 This can be verified by examining
𝑥→2
the function’s graph and table of values near x = 2.
These well-behaved functions are considered continuous at c. We will
examine this concept more in a later lesson.
Now let’s look at some properties of limits that will help us when
performing direct substitution.
Properties of Limits
If L , M , c , and k are real num bers and
lim f
x c
1.
x L
Sum R ule :
lim g  x   M , then
and
lim  f
x c
x c
 x   g  x   L  M
T he lim it of the sum of tw o functions is the sum of their lim its.
2.
D ifferenceR ule :
lim  f
x c
 x   g  x   L  M
T he lim it of the difference of tw o functions is t he difference
of their lim its.
Properties of Limits continued
Product Rule:
Constant Multiple Rule:
Properties of Limits continued
6.
If r and s are integers, s  0, then
P ow er R ule :
r
r
lim  f  x   s  L s
x c
r
provided that L s is a real num ber.
T he lim it of a rational pow er of a funct ion is that pow er of the
lim it of the function, provided the latt e r is a real num ber.
O ther properties of lim its:
lim  k   k
x c
lim  x   c
x c
Polynomial and Rational Functions
Trying to evaluate functions using just these rules can be a grueling process (I’m not going to
make you do that) so it is a little easier to just combine a bunch of those rules to evaluate
polynomial and rational functions.
1.
If f  x   a n x  a n 1 x
n
n 1
 ...  a 0 is any polynom ial function and
c is any real num ber, then
lim f  x   f  c   a n c  a n 1 c
n
x c
2.
n 1
 ...  a 0
If f  x  and g  x  are polynom ials and c is any real num ber, then
lim
x c
f x
g x

f c
g c
, prov ided that g  c   0.
Now let’s look at a few examples using these new properties and rules.
Example Limits
Use the rules from the previous slides to find the following limits using direct substitution.
a) lim (4𝑥 2 − 2𝑥 + 6)
𝑥→5
lim  4 x - 2 x  6   4  5   2  5   6  4  25   1 0  6  100  1 0  6  9 6
2
2
x5
b)
𝑥 2 +𝑥+2
lim
𝑥→1 𝑥+1
𝑥 2 +𝑥+2
lim
𝑥→1 𝑥+1
=
12 +1+2
1+1
=
4
2
=
2
Evaluating Limits
As with polynomials, limits of many familiar functions can be found by
substitution at points where they are defined. This includes trigonometric,
exponential, and logarithmic functions, and composites of these functions.
Now lets try an example:
F in d
lim
x 0
1  sin x
co s x
SLet’s
olve look
graphically:
at the graph first:
T he graph of f
x
1  sin x
suggests that the lim it e xists and is 1.
C oxn firm
cos
If we can, we always want
F in d
to confirm this algebraically:
A n alytically:
lim
x 0
1  sin x

lim  1  sin x 
x 0
co s x
lim co s x
x 0

1 0
1
1

1  sin 0 
co s 0
Evaluating Limits cont.
Now let’s try another example:
F ind lim
x 0
5
x
C onfirm A nalytically :
W e can't use substitution in this exam ple because w hen x is relaced by 0,
the denom inator becom es 0 and the function is undefined.
T his w ould suggest that w e rely on the g raph to s ee that the
lim it does not exist.
Dividing Out Technique
Sometimes direct substitution fails even when the limit exists. This is one
technique to help find the limit in that case. Let’s use it in an example
Find:
𝑥 2 +𝑥−6
lim
𝑥→−3 𝑥+3
In this case here we can’t use direct substitution because if
0
we try to we will get . Let’s see what the graph looks like.
0
One approach we can try to do is factor our original function
and see if we can cancel anything out that may allow us to do
direct substitution.
𝑥 2 +𝑥−6
𝑥+3
There appears to be a limit as x → −3
=
(𝑥+3)(𝑥−2)
(𝑥+3)
=𝑥−2
(𝑥 ≠ −3)
Now we can easily find the limit using direct substitution
lim (𝑥 − 2) = −5
𝑥→−3
Rationalizing Technique
Another technique that we can try when direct substitution fails but a limit
appears to exist is to rationalize the numerator or denominator and see if that
helps us to make a cancellation that allows for us to use direct substitution.
𝑥+1−1
𝑥
𝑥→0
Find lim
Now just like in the last problem, direct substitution won’t
0
work because we get the form . If you look at the graph, it appears that there is a limit as
0
x approaches 1. There is no dividing out that we can do, so let’s try rationalizing the
numerator and see what happens.
𝑥+1−1
=
𝑥
1
𝑥+1+1
𝑥+1−1
𝑥
𝑥+1+1
𝑥+1+1
=
𝑥+1 −1
𝑥( 𝑥 + 1 + 1)
=
𝑥
𝑥( 𝑥 + 1 + 1)
1
1
Now we can easily find the limit lim
=
𝑥→0
𝑥+1+1 2
using direct substitution.
=
Recap – Finding Limits Algebraically
A few important points on the last two examples that we just did.
• What we are actually doing when we divide out or rationalize is find a new function. This
function is technically different from our original function because it has a different domain.
But what you will notice if you look at the graphs or tables of values of our original and new
function is that they have the exact same values at every point EXCEPT at the point(s) where
our original function is undefined. While in practice this doesn’t change your approach to the
problem, it is important to note that while the functions aren’t technically equal at EVERY
point (because the original function is undefined at some point(s)), their LIMITS ARE EQUAL
AT EVERY POINT.
0
,
0
• As a general rule, when you try direct substitution and you get this is called indeterminate
form and a limit exists at that point. You should then try rationalizing or dividing out to make
the denominator ≠ 0 and then try direct substitution again.
• If you try direct substitution and you get
NOT exist at that point.
𝑐
0
where c is a non-zero constant, then a limit will
A General Strategy for Finding Limits
1) Try direct substitution. Direct substitution will work for most types of
functions. For piecewise functions, if you are looking for the limit at an x
value where the rule for the function changes, use direct substitution on the
left and right rules at that point and see if they are equal. Remember, in order
for a limit to exist, both of the one-sided limits must be equal to each other.
2) If you can’t evaluate the limit of f(x) at a point c using direct substitution,
try dividing out or rationalizing to find a new function g that agrees with f for
all x other than x = c. Choose a g where you can find the limit at x = c using
direct substitution.
3) Always confirm or reinforce your conclusion using a graph or table of values
or both if you can. It doesn’t matter if you check before or after you try direct
substitution, but it’s very important to make sure that your answer makes
sense. If you don’t have calculator access, you just have to trust your algebra.
The Sandwich (or Squeeze) Theorem
If w e cannot find a lim it directly, w e m ay be able to find
it indirectly w ith the S andw ich T heorem . T he theorem
refers to a function f w hose values are s andw iched betw een
the values of tw o other func tions, g and h .
If g and h have the sam e lim it as x  c then f has that lim it too.
If g  x   f  x   h  x  for all x  c in som e interval about c , an d
lim g  x  = lim h  x  = L ,
x c
x c
then
lim f  x  = L
x c
Find
1
𝑥
lim 𝑥 2 sin
𝑥→0
Now we know that sin
1
𝑥
Sandwich Theorem Example
In this case here, we can’t use direct substitution and we can’t
divide out or rationalize anything to help with that sin
≤ 1 because of the range of sine. Therefore
We also know that -1 ≤ sin
1
𝑥
1
𝑥
≤ 𝑥 2· 1
again because of the range of sine. Therefore 𝑥 2 · (-1) ≤ 𝑥 2 sin
-𝑥 2 ≤ 𝑥 2 sin
This leads to the following:
𝑥 2 sin
1
𝑥
1
𝑥
1
𝑥
≤ 𝑥2
This means that our function is sandwiched between -𝑥 2 and 𝑥 2 for all values of 𝑥.
Let’s try some limits:
lim (-𝑥 2 ) ≤ lim 𝑥 2 sin
1
𝑥
≤ lim (𝑥 2 )
0 ≤ lim 𝑥 2 sin
1
𝑥
≤0
𝑥→0
By direct substitution:
𝑥→0
𝑥→0
𝑥→0
which means that the only possible answer is 0!
Take a look at a graph of the three functions we just worked with in this example.
Notice that the function with the sin
1
in it is “sandwiched” between the other functions for all x values.
.
Two Special Trigonometric Limits
There are two special trig limits that you need to memorize. The first one is much more likely
to appear on tests in this class and on the AP exam but they are both worth memorizing.
sin 𝑥
𝑥→0 𝑥
𝟏. lim
=1
1−cos 𝑥
𝑥
𝑥→0
𝟐. lim
=0
On your own, you should verify both of these graphically and/or numerically.
Example With a Piecewise Function
4𝑥 − 1,
Find lim 𝑓(𝑥) where 𝑓 𝑥 = 2
𝑥→2
𝑥 + 3,
𝑥≤2
𝑥>2
Remember that we can only say that a limit exists at a given x value if both of the one-sided limits exist and are
equal at that x value. In this case, we will use direct substitution in both of the given definitions of f(x) since x=2 is
the value where the definition of f(x) changes.
lim− 𝑓 𝑥 = 4 2 − 1 = 7
𝑥→2
Since both one sided limits are equal at x = 2, we can say that
lim+ 𝑓(𝑥) = 22 + 3 = 7
𝑥→2
lim 𝑓 𝑥 = 7.
𝑥→2
In symbols we would say:
lim 𝑓 𝑥 = lim+ 𝑓(𝑥) = 7
𝑥→2−
𝑥→2
therefore
lim 𝑓 𝑥 = 7
𝑥→2
The symbols representation in the box above is what is expected of you on my exams and on
the AP exam for Free-Response questions as justification.
Summary
• For most functions where the denominator ≠ 0 at the x value where you are trying to find a
limit, direct substitution is the most accurate way to find a limit. The properties of limits will
also help to simplify more difficult limits with direct substitution.
0
• If direct substitution gives an answer of the form then try the methods of factoring out or
0
rationalizing to find a new function where direct substitution will work.
• If none of the above methods work, the Sandwich Theorem may help.
𝑐
• If direct substitution gives an answer of the form where c is a non-zero constant, then the
0
limit will not exist at that point.
• For piecewise functions, be sure that both one-sided limits are equal at the point where you
are trying to find the limit if it is at an x value where the rules of the function change.
• Memorize and be able to use the following two special properties:
sin 𝑥
𝑥→0 𝑥
𝟏. lim
=1
1−cos 𝑥
𝑥
𝑥→0
𝟐. lim
=0