limits - mathematics department : welcome · · 2016-06-27limits from last time... ... and...
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Limits
From last time...
Let y = f (t) be a function that gives the position at time t of anobject moving along the y -axis. Then
Ave vel[t1
, t2
] =f (t
2
)� f (t1
)
t
2
� t
1
Velocity(t) = limh!0
f (t + h)� f (t)
h
.
a a+h
f(a)
f(a+h)
h
We need to be able to take limits!
Limit of a Function – Definition
We say that a function f approaches the limit L as x approaches a,
written limx!a
f (x) = L,
if we can make f (x) as close to L as we want by taking x
su�ciently close to a.
a
L
x
f(x)
Δx
Δy
i.e. If you need �y to be smaller,you only need to make �x smaller(� means ”change”)
One-sided limits
Lr
Ll
a
f(a)
Right-handed limit: Lr
= limx!a
+
f (x)
if f (x) gets closer to L
r
as x gets closer to a from the right
Left-handed limit: L` = limx!a
�f (x)
if f (x) gets closer to L` as x gets closer to a from the left
Theorem
The limit of f as x ! a exists if and only if both the right-hand
and left-hand limits exist and have the same value, i.e.
limx!a
f (x) = L if and only if limx!a
�f (x) = L and lim
x!a
+
f (x) = L.
Examples
limx!2
x � 2
x + 3
= 0
limx!1
x
2 � 1
x � 1
= 2
limx!0
1
x
is undefined
-4 4
-8
-4
4
8
-4 4
-8
-4
4
8
-4 4
-8
-4
4
8
Theorem
If limx!a
f (x) = A and limx!a
g(x) = B both exist, then
1. limx!a
(f (x) + g(x)) = limx!a
f (x) + limx!a
g(x) = A+ B
2. limx!a
(f (x)� g(x)) = limx!a
f (x)� limx!a
g(x) = A� B
3. limx!a
(f (x)g(x)) = limx!a
f (x) · limx!a
g(x) = A · B4. lim
x!a
(f (x)/g(x)) = limx!a
f (x)/ limx!a
g(x) = A/B (B 6=0).
In short: to take a limit
Step 1: Can you just plug in? If so, do it.
Step 2: If not, is there some sort of algebraic manipulation (likecancellation) that can be done to fix the problem? If so, do it.Then plug in.
Step 3: Learn some special limit to fix common problems. (Later)
If in doubt, graph it!
Examples
1. limx!2
x � 2
x + 3= 0 because if f (x) = x�2
x+3
, then f (2) = 0.
2. limx!1
x
2 � 1
x � 1!0
!0
If f (x) = x
2�1
x�1
, then f (x) is undefined at x = 1.However, so long as x 6= 1,
f (x) =x
2 � 1
x � 1=
(x + 1)(x � 1)
x � 1= x + 1.
So
limx!�1
x
2 � 1
x � 1= lim
x!1
x + 1 = 1 + 1 = 2 .
3. limx!0
px + 2�
p2
x
!0
!0
, so again, f (x) is undefined at a.
Multiply top and bottom by the conjugate:
limx!0
px + 2�
p2
x
= limx!0
px + 2�
p2
x
! px + 2 +
p2
px + 2 +
p2
!
= limx!0
x + 2� 2
x(px + 2 +
p2)
since (a � b)(a + b) = a
2 � b
2
= limx!0
x
x(px + 2 +
p2)
= limx!0
1px + 2 +
p2= 1
2
p2
Examples
3. limx!0
px + 2�
p2
x
!0
!0
, so again, f (x) is undefined at a.
Multiply top and bottom by the conjugate:
limx!0
px + 2�
p2
x
= limx!0
px + 2�
p2
x
! px + 2 +
p2
px + 2 +
p2
!
= limx!0
x + 2� 2
x(px + 2 +
p2)
since (a � b)(a + b) = a
2 � b
2
= limx!0
x
x(px + 2 +
p2)
= limx!0
1px + 2 +
p2= 1
2
p2
Examples
1. limx!1
x
2�3x+2
x
2
+4x�5
2. limx!�2
|x |x
= �1
3. limx!0
|x |x
is undefined
4. limx!0
(3 + x)2 � 32
x
= �1
6
= 6
Infinite limitsIf f (x) gets arbitrarily large as x ! a, then it doesn’t have a limit.Sometimes, though, it’s more useful to give more information.
Example:
For both f (x) = 1
x
and f (x) = 1
x
2
, limx!0
f (x) does not exist.However, they’re both “better behaved” than that might imply:
limx!0
+
1
x
= 1, limx!0
�
1
x
= �1 limx!0
+
1
x
2
= 1, limx!0
�
1
x
2
= 1
limx!0
1
x
does not exist limx!0
1
x
2
= 1
Why? A vertical asymptote occurs where
limx!a
+
f (x) = ±1 and limx!a
�f (x) = ±1
Infinite limitsBadly behaved example:
f (x) = csc(1/x)
-0.5 0.5
Zoom way in:
-0.005 0.005
(denser and denser vertical asymptotes)
limx!0
+
csc(1/x) does not exist, and limx!0
�csc(1/x) does not exist
Limits at InfinityWe say that a function f approaches the limit L as x gets biggerand bigger (in the positive or negative direction),
written limx!1
f (x) = L or limx!�1
f (x) = L
if we can make f (x) as close to L as we want by taking x
su�ciently large. (By “large”, we mean large in magnitude)
Example 1:
limx!�1
1
x
= 0 and limx!1
1
x
= 0.
Limits at Infinity: functions and their inverses
Theorem
If limx!a
± f (x) = 1, then limx!1 f
�1(x) = a.
If limx!a
± f (x) = �1, then limx!�1 f
�1(x) = a.
Example: Let arctan(x) be the inverse function to tan(x):y = tan(x):
π/2-π/2
Since limx!⇡/2� = 1
and limx!�⇡/2+ = �1
(restrict the domain to (�⇡2
, ⇡2
) to that we can invert)
y = arctan(x):
-π/2
π/2
we have limx!1 = ⇡/2
and limx!�1 = �⇡/2
Rational functionsLimits that look like they’re going to 1
1 can actually be doing lotsof di↵erent things. To fix this, divide the top and bottom by thehighest power in the denominator!
Ex 1 limx!1
x � 1
x
3 + 2!1!1
Fix: multiply the expression by 1/x3
1/x3 :
limx!1
x � 1
x
3 + 2= lim
x!1
✓x � 1
x
3 + 2
◆✓1/x3
1/x3
◆= lim
x!1
1
x
2
� 1
x
3
1 + 2
x
3
=0� 0
1 + 0= 0
Ex 2 limx!1
3x2 � 2x + 1
4x2 � 1!1!1
Fix: multiply the expression by 1/x2
1/x2
Ex 3 limx!1
x
4 � x
2 + 2
x
3 + 3!1!1
Fix: multiply the expression by 1/x3
1/x3
Rational functions: quick trick
limx!1
x � 1
x
3 + 2= 0 lim
x!1
3x2 � 2x + 1
4x2 � 1=
3
4lim
x!1
x
4 � x
2 + 2
x
3 + 3= 1
Suppose
P(x) = a
n
x
n+· · ·+a
1
x+a
0
and Q(x) = b
m
x
m+· · ·+b
1
x+b
0
are polynomials of degree n and m respectively. Then in general,
limx!1
P(x)
Q(x)= lim
x!1
a
n
x
n
b
m
x
m
= limx!1
a
n
b
m
x
n�m =
8><
>:
0 n < m
a
n
b
m
n = m
±1 n < m
Examples: Other ratios with powers.Example:
limx!1
xp3x2 + 2
[hint: multiply by 1/x1/x and remember a
pb =
pa
2
b.]
limx!1
xp3x2 + 2
= limx!1
✓xp
3x2 + 2
◆✓1/x
1/x
◆
= limx!1
11
x
p3x2 + 2
= limx!1
1q1
x
2
(3x2 + 2)
= limx!1
1q3 + 2
x
2
=1p3 + 0
= 1p3
Evaluating limits when x ! 0.
1. Show lim
x!0(x
2 � 2)
2+ 6 = 10.
2. Show lim
x!0
5x
x
= 5.
3. Show lim
x!0
17x
2x
= 17/2.
4. Show lim
x!0
�317x
422x
= �317/422.
5. Show lim
x!0
�317x� 3
422x+ 5
= �3/5.
6. Show lim
h!0
px+ h�
px
h
=
1
2
px
.
7. Show lim
x!0
p1 + x+ x
2 � 1
x
= 1/2.
8. Show lim
x!0
p3 + x�
p3
x
= 1/(2
p3).
9. Show lim
h!0
1
h
✓1p
x+ h
� 1px
◆= �(1/2)x
�3/2.
10. Show lim
x!0
2xpa+ x�
pa� x
= 2
pa.
11. Show lim
x!0
p1 + x� 1
x
= 1/2.
12. Show lim
x!0
xp1 + x� 1
= 2.
13. Show lim
x!0
e
x
+ e
�x � 2
x
2= 1.
14. Show
lim
h!0
f(x+ h)� f(x)
h
=
a
2
pax+ b
when f(x) =
pax+ b.
15. Show
lim
h!0
f(x+ h)� f(x)
h
= mn(mx+ c)
n�1
when f(x) = (mx+ c)
n
.
Evaluating limits when x! a.
1. Show lim
x!1(6x
2 � 4x + 3) = 5.
2. Show lim
x!7
x
2 � 49
x� 7
= 14.
3. Show lim
x!2
x
2 � 6x + 8
x� 2
= �2.
4. Show lim
x!�5
2x
2+ 9x� 5
x + 5
= �11.
5. Show lim
x!1
x
3 � 1
x� 1
= 3.
6. Show lim
x!3
x
2 � 4x + 3
x
2 � 2x� 3
= 1/2.
7. Show lim
x!�2
x
3+ 8
x + 2
= 4.
8. Show lim
x!3
x
4 � 81
x� 3
= 108.
9. Show lim
x!5
x
5 � 3125
x� 5
= 3125.
10. Show lim
x!a
x
12 � a
12
x� a
= 12a
11.
11. Show lim
x!a
x
5/2 � a
5/2
x� a
= (5/2)a
3/2.
12. Show lim
x!a
(x + 2)
5/3 � (a + 2)
5/3
x� a
= (5/3)(a + 2)
2/3.
13. Show lim
x!4
x
3 � 64
x
2 � 16
= 6.
14. Show lim
x!2
x
5 � 32
x
3 � 8
= 20/3.
15. Show lim
x!1
x
n � 1
x� 1
= n.
16. Show lim
x!a
px�
pa
x� a
=
1
2
pa
.
17. Show lim
x!2
p3� x� 1
2� x
= 1/2.
18. Show lim
x!a
pa + 2x�
p3xp
3a + x� 2
px
=
2
p3
9
.
19. Show lim
x!a
x
n � a
n
x� a
= na
n�1.
Evaluating limits when x!1.
1. Show lim
x!1
x + 2
x� 2
= 1.
2. Show lim
x!1
3x
2+ 2x� 5
5x
2+ 3x + 1
= 3/5.
3. Show lim
x!1
x
2 � 7x + 11
3x
2+ 10
= 1/3.
4. Show lim
x!1
2x
3 � 5x + 7
7x
3+ 2x
2 � 6
= 2/7.
5. Show lim
x!1
(3x� 1)(4x� 5)
(x + 6)(x� 3)
= 12.
6. Show lim
x!1
xp4x
2+ 1� 1
= 1/2.
7. Show lim
x!�12
x
= 0.
8. Show lim
t!1
t + 1
t
2+ 1
= 0.
9. Show lim
n!1
pn
2+ 1� n = 0.
10. Show lim
n!1
pn
2+ n� n = 1/2.