Math 2413
1. (2 pts)
(a)
=
f ( −1) 2
Homework Set 2 – Solutions
10 Points
=
lim− f ( x ) 2 =
lim+ f ( x ) 1
x →−1
x →−1
=
lim f ( x ) doesn't exist b/c lim− f ( x ) ≠ lim+ f ( x )
x →−1
x →−1
x →−1
−3
lim− f ( x ) =
−1
lim+ f ( x ) =
−1
lim f ( x ) =
−1
(b) f ( 2 ) =
x→2
x→2
x→2
=
lim− f ( x ) 4=
lim+ f ( x ) 4 =
lim f ( x ) 4
(c) f ( 3) doesn't exist=
x →3
3. (2 pts) lim
x →3
x →3
x →3
( x − 3)( x + 2 ) = lim x + 2 = 5 = − 5
x2 − x − 6
= lim
2
x →3
x →3 − x
3x − x
− x ( x − 3)
−3
3
5. (2 pts)
(
(
)
)
25 − ( 4 − 3 y )
5 − 4 − 3y 5 + 4 − 3y
21 + 3 y
= lim
lim = lim
y →−7
y →−7
y+7
5 + 4 − 3y
( y + 7 ) 5 + 4 − 3 y y →−7 ( y + 7 ) 5 + 4 − 3 y
(
(
)
)
3
3
= lim
=
y →−7 5 + 4 − 3 y
10
6. (2 pts for (b) ONLY – part (a) not graded)
(a) In this case we can use the second formula because x= 12 > −3 is completely inside this region.
lim g ( x=
) lim ( x + 7=) 19
x → 12
x →12
(b) Here we will need to look at the two one-sided limits because x = −3 is the “cut-off” point.
lim
=
g ( x)
−
x →−3
lim+ g (=
x)
x →−3
lim
=
e 2 − x e5
−
lim g ( x ) ≠ lim+ g ( x )
x →−3
lim+ ( x +=
7) 4
x →−3−
x →−3
x →−3
The two one-sided limits are not the same and so lim g ( x ) doesn’t exist.
x→ 6
8. (2 pts) In both of these limits the numerator is staying fixed at 9 and as x approaches 3 (from either
side) we can see that 6-2x is approaching zero. So, we have a fixed number divided by something
increasingly smaller and so it should make some sense that both of these are going to either ∞ or −∞
and this will depend upon the sign of the denominator
In the first case 6-2x is positive because x < 3 and raising this to the 5th power will not change this. So,
in this case we have a fixed positive number in the numerator divided by something increasingly smaller
and positive and so the limit in this case will be ∞ .
Math 2413
Homework Set 2 – Solutions
10 Points
In the second case 6-2x is negative because x > 3 and raising this to the 5th power will not change this.
So, in this case we have a fixed positive number in the numerator divided by an increasingly smaller and
negative number and so the limit in this case will be −∞ .
Because the two one-sided limits are not the same the overall limit does not exist. Here are the official
answers to this problem.
lim
x →3 −
9
(6 − 2x)
5
= ∞
lim
x →3 +
9
(6 − 2x)
5
= −∞
lim
x →3
9
(6 − 2x)
5
doesn't exist
Not Graded
2. I believe I used the same property numbers in class as in my notes online, but just in case the
properties listed here are those from the notes online.
(a)
− 9  lim f ( x ) + lim 7 g ( x ) − lim 2
lim  f ( x ) + 7 g ( x )=
Property 2
=lim f ( x ) + 7 lim g ( x ) − lim 2
Property 1
=lim f ( x ) + 7 lim g ( x ) − 2
Property 7
= 9 + 7 ( −1) − 2 = 0
Plug in values of limits
x →12
x →12
x →12
x →12
x →12
x →12
(b)
lim
x →12
g ( x)
=
 f ( x )  − 5h ( x )
2
=
=
=
4. lim
t →1
( t − 1)( t − 4 ) + 5 − 5t =
4t − 3 − t 2
(
x →12
x →12
x →12
lim g ( x )
x →12
lim  f ( x )  − 5h ( x )
x →12 
2
)
Property 4
lim g ( x )
x →12
lim  f ( x )  − lim 5h ( x )
x →12 
x →12
2
Property 2
lim g ( x )
x →12
2
 lim f ( x )  − 5 lim h ( x )

x →12
 x →12
−1
[9]
2
lim
t →1
− 5 ( −4 )
= −
1
101
Property 5 & 1
Plug in values of limits
( t − 9 )( t − 1) = lim t − 9 = −8 = −4
t 2 − 10t + 9
= lim
2
− ( t − 4t + 3) t →1 − ( t − 3)( t − 1) t →1 − ( t − 3) − ( −2 )
Math 2413
Homework Set 2 – Solutions
10 Points
7. Note that we can’t just cancel the h’s since once is inside the absolute value bars. So, we’ll use the
hint and recall that,
if h ≥ 0
h
h =
−h if h < 0
With this we can do the two one sided limits to eliminate the absolute value bars.
h
−h
=lim−
=lim − 1 =−1
h →0 h
h →0
h h →0 −
h
h
=
=
lim=
lim
lim
1 1
+
+
h →0 h
h →0 h
h →0 −
because h < 0 in this case
lim−
because h > 0 in this case
The two one-sided limits are not the same and so lim
h →0
h
does not exist.
h
9. Note : There was a typo in this problem. The functions should have been w + 9
In both of these limits the numerator is staying fixed at -2 and as w approaches -9 (from either side) we
can see that w+9 is approaching zero. So, we have a fixed number divided by something increasingly
smaller and so it should make some sense that both of these are going to either ∞ or −∞ and this will
depend upon the sign of the denominator
In the first case w+9 is negative because w < −9 and raising this to the 10th power will make this
positive. So, in this case we have a fixed negative number in the numerator divided by something
increasingly smaller and positive and so the limit in this case will be −∞ .
In the second case w+9 is positive because w > −9 and raising this to the 10th power will not change
this. So, in this case we have a fixed negative number in the numerator divided by an increasingly
smaller and positive number and so the limit in this case will be −∞ .
Alternatively instead of the previous two paragraphs we could just acknowledge that the exponent in
the denominator is even and so the denominator will be positive regardless of the sign of w+9.
Because the two one-sided limits are the same the overall limit will also be −∞ . Here are the official
answers to this problem.
lim −
w→−9
−2
( w + 9)
10
= −∞
lim +
w→−9
−2
( w + 9)
10
= −∞
lim
w→−9
−2
( w + 9)
10
= −∞