Lecture 4 : Calculating Limits using Limit Laws
Lecture 4 : Calculating Limits using Limit Laws
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Using the definition of the limit, limx¡úa f (x), we can derive many general laws of limits, that help us to
calculate limits quickly and easily. The following rules apply to any functions f (x) and g(x) and also
apply to left and right sided limits:
Suppose that c is a constant and the limits
lim f (x)
x¡úa
exist (meaning they are finite numbers).
and
lim g(x)
x¡úa
Then
1. limx¡úa [f (x) + g(x)] = limx¡úa f (x) + limx¡úa g(x) ;
(the limit of a sum is the sum of the limits).
2. limx¡úa [f (x) ? g(x)] = limx¡úa f (x) ? limx¡úa g(x) ;
(the limit of a difference is the difference of the limits).
3. limx¡úa [cf (x)] = c limx¡úa f (x);
(the limit of a constant times a function is the constant times the limit of the function).
4. limx¡úa [f (x)g(x)] = limx¡úa f (x) ¡¤ limx¡úa g(x);
(The limit of a product is the product of the limits).
(x)
x¡úa f (x)
= lim
if limx¡úa g(x) 6= 0;
5. limx¡úa fg(x)
limx¡úa g(x)
(the limit of a quotient is the quotient of the limits provided that the limit of the denominator is
not 0)
Example
If I am given that
lim f (x) = 2,
x¡ú2
lim g(x) = 5,
x¡ú2
lim h(x) = 0.
x¡ú2
find the limits that exist (are a finite number):
2f (x) + h(x) limx¡ú2 (2f (x) + h(x))
=
since lim g(x) 6= 0
x¡ú2
x¡ú2
g(x)
limx¡ú2 g(x)
(a) lim
=
2 limx¡ú2 f (x) + limx¡ú2 h(x)
2(2) + 0
4
=
=
limx¡ú2 g(x)
5
5
f (x)
x¡ú2 h(x)
f (x)h(x)
x¡ú2
g(x)
(b) lim
(c) lim
Note 1 If limx¡úa g(x) = 0 and limx¡úa f (x) = b, where b is a finite number with b 6= 0, Then:
(x)
the values of the quotient fg(x)
can be made arbitrarily large in absolute value as x ¡ú a and thus
1
the limit does not exist.
(x)
If the values of fg(x)
are positive as x ¡ú a in the above situation, then limx¡úa
f (x)
= ¡Þ,
g(x)
(x)
limx¡úa fg(x)
= ?¡Þ,
(x)
If the values of fg(x)
are negative as x ¡ú a in the above situation, then
If on the other hand, if limx¡úa g(x) = 0 = limx¡úa f (x), we cannot make any conclusions about
the limit.
Example Find limx¡ú¦Ð?
cos x
.
x?¦Ð
As x approaches ¦Ð from the left, cos x approaches a finite number ?1.
As x approaches ¦Ð from the left, x ? ¦Ð approaches 0.
Therefore as x approaches ¦Ð from the left, the quotient
cos x
x?¦Ð
approaches ¡Þ in absolute value.
The values of both cos x and x ? ¦Ð are negative as x approaches ¦Ð from the left, therefore
lim?
x¡ú¦Ð
cos x
= ¡Þ.
x?¦Ð
More powerful laws of limits can be derived using the above laws 1-5 and our knowledge of some
basic functions. The following can be proven reasonably easily ( we are still assuming that c is a
constant and limx¡úa f (x) exists );
n
6. limx¡úa [f (x)]n = limx¡úa f (x) , where n is a positive integer (we see this using rule 4 repeatedly).
7. limx¡úa c = c, where c is a constant ( easy to prove from definition of limit and easy to see from
the graph, y = c).
8. limx¡úa x = a, (follows easily from the definition of limit)
9. limx¡úa xn = an where n is a positive integer (this follows from rules 6 and 8).
¡Ì
¡Ì
10. limx¡úa n x = n a, where n is a positive integer and a > 0 if n is even. (proof needs a little extra
work and the binomial theorem)
p
p
11. limx¡úa n f (x) = n limx¡úa f (x) assuming that the limx¡úa f (x) > 0 if n is even. (We will look at
this in more detail when we get to continuity)
Example Evaluate the following limits and justify each step:
(a)
limx¡ú3
(b)
limx¡ú1
x3 +2x2 ?x+1
x?1
¡Ì
3
x+1
2
(c)
Determine the infinite limit (see note 1 above, say if the limit is ¡Þ, ?¡Þ or D.N.E.)
x+1
limx¡ú2? (x?2)
.
Polynomial and Rational Functions
Please review the relevant parts of Lectures 3, 4 and 7 from the Algebra/Precalculus review
page. This demonstration will help you visualize some rational functions:
Direct Substitution (Evaluation) Property If f is a polynomial or a rational function and
a is in the domain of f , then limx¡úa f (x) = f (a). This follows easily from the rules shown above.
(Note that this is the case in part (a) of the example above)
if f (x) =
P (x)
Q(x)
is a rational function where P (x) and Q(x) are polynomials with Q(a) = 0, then:
P (x)
If P (a) 6= 0, we see from note 1 above that limx¡úa Q(x)
= ¡À¡Þ or D.N.E. and is not equal to ¡À¡Þ.
If P (a) = 0 we can cancel a factor of the polynomial P (x) with a factor of the polynomial Q(x)
and the resulting rational function may have a finite limit or an infinite limit or no limit at x = a.
P (x)
by the following observation which
The limit of the new quotient as x ¡ú a is equal to limx¡úa Q(x)
we made in the last lecture:
Note 2: If h(x) = g(x) when x 6= a, then limx¡úa h(x) = limx¡úa g(x) provided the limits exist.
Example
to ¡À¡Þ:
Determine if the following limits are finite, equal to ¡À¡Þ or D.N.E. and are not equal
(a) limx¡ú3
x2 ?9
.
x?3
(b)
limx¡ú1?
x2 ?x?6
.
x?1
(c) Which of the following is true:
2 ?x?6
1. limx¡ú1 x x?1
= +¡Þ,
2. limx¡ú1
¡À¡Þ,
x2 ?x?6
x?1
3
= ?¡Þ,
3. limx¡ú1
x2 ?x?6
x?1
D.N.E. and is not
Example
Evaluate the limit (finish the calculation)
(3 + h)2 ? (3)2
.
h¡ú0
h
lim
limh¡ú0
(3+h)2 ?(3)2
h
Example
= limh¡ú0
9+6h+h2 ?9
h
=
Evaluate the following limit:
¡Ì
lim
x¡ú0
x2 + 25 ? 5
.
x2
Recall also our observation from the last day which can be proven rigorously from the definition
(this is good to keep in mind when dealing with piecewise defined functions):
Theorm limx¡úa f (x) = L
if and only if
limx¡úa? f (x) = L = limx¡úa+ f (x).
Example Evaluate the limit if it exists:
3x + 6
x¡ú?2 |x + 2|
lim
The following theorems help us calculate some important limits by comparing the behavior of a
function with that of other functions for which we can calculate limits:
4
Theorem If f (x) ¡Ü g(x) when x is near a(except possible at a) and the limits of f (x) and g(x)
both exist as x approaches a, then
lim f (x) ¡Ü lim g(x).
x¡úa
x¡úa
The Sandwich (squeeze) Theorem
If f (x) ¡Ü g(x) ¡Ü h(x) when x is near a (except
possibly at a) and
lim f (x) = lim h(x) = L
x¡úa
x¡úa
then
lim g(x) = L.
x¡úa
Recall last day, we saw that limx¡ú0 sin(1/x) does not exist because of how the function oscillates near x = 0. However we can see from the graph below and the above theorem that
limx¡ú0 x2 sin(1/x) = 0, since the graph of the function is sandwiched between y = ?x2 and
1
y = x2 :
0.5
K1
0
K0.5
0.5
x
1
K0.5
Example Calculate the limit limx¡ú0 x2 sin x1 . K1
We have ?1 ¡Ü sin(1/x) ¡Ü 1O for all x,
multiplying across by x2 (which is positive), we get ?x2 ¡Ü x2 sin(1/x) ¡Ü x2 for all x,
Using the Sandwich theorem, we get
0 = lim ?x2 ¡Ü lim x2 sin(1/x) ¡Ü lim x2 = 0
x¡ú0
x¡úo
x¡ú0
Hence we can conclude that
lim x2 sin(1/x) = 0.
x¡ú0
Example
Decide if the following limit exists and if so find its values:
limx¡úo x100 cos2 (¦Ð/x)
5
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