Math 2413
Homework Set 2
1. (2 pts)
−2
(a) f ( −3) =
lim f ( x ) =
2
x →−3−
lim f ( x ) =
2
10 Points
lim f ( x ) =
2
x →−3+
x →−3
=
lim− f ( x ) 0=
lim+ f ( x ) 0 =
lim f ( x ) 0
(b) f ( 0 ) 0 =
x →0
x →0
x →0
d.n.e
lim− f ( x ) =
lim+ f ( x ) =
lim f ( x ) =
−4
−4
−4
(c) f ( 2 ) =
x→2
x→2
x→2
(d)
lim− f ( x ) 2
f ( 3) 2=
=
x →3
lim f ( x )
=
x →3
lim+ f ( x ) 4
=
x →3
doesn't exist b/c lim− f ( x ) ≠ lim+ f ( x )
x →−3
x →−3
− ( t − 5 )( t + 5 )
− ( t − 5)
25 − t 2
10
10
= lim
= lim
=
= −
3. (2 pts) lim 2
t →−5 2t + 9t − 5
t →−5 ( 2t − 1)( t + 5 )
t →−5 2t − 1
11
−11
5. (2 pts)
y2 + 5 − 3
y2 + 5 − 3 y2 + 5 + 3
y2 + 5 − 9
lim 2
lim
lim
=
=
y →2 y + 2 y − 8
y →2 y 2 + 2 y − 8
y 2 + 5 + 3 y →2 ( y 2 + 2 y − 8 ) y 2 + 5 + 3
(
( y − 2 )( y + 2 ) =
y →2
( y + 4 )( y − 2 ) ( y 2 + 5 + 3)
= lim
lim
y →2
y+2
( y + 4) (
)
=
y +5 +3
2
)
4
=
( 6 )( 6 )
1
9
6. (2 pts – Part b ONLY) For (a) we can just use the lower formula since 11 ≥ 6 contains points on both
sides in this interval. For (b) we’ll need to look at the two one-sided limits since x = 6 is the cutoff
point and no interval contains points on both sides of it.
(
)
(a) lim g (=
x ) lim x 2 −=
3 118
x →11
x →11
(b) First the one sided limits.
lim
=
g ( x ) lim
=
e 2 + x e8
−
x → 6−
x →6
x → 6+
x →6
lim g =
3) 33
( x ) lim+ ( x 2 −=
because x < 6 in this case .
because x > 6 in this case
So, lim g ( x ) doesn’t exist since the two one-sided limits are not the same.
x →1
9. (2 pts) In both of these limits the numerator is staying fixed at -5 and as x approaches -3 (from either
side) we can see that x+3 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 −∞ .
Also note that in this case because we’re squaring the denominator it will always be positive and so both
of these limits will have a fixed negative numerator divided by an increasingly small positive number.
Therefore, both will be −∞ .
Math 2413
Homework Set 2
10 Points
Here are the official answers to this problem.
lim−
t →−3
−5
( 2x + 6)
2
= −∞
lim+
t →−3
−5
( 2x + 6)
2
= −∞
Not Graded
2.
(a)
lim 9 g ( x ) − 2h ( x )=
9 g ( x ) − lim 2h ( x )
 xlim
→− 2
x →− 2
Property 2
= 9 lim g ( x ) − 2 lim h ( x )
Property 1
x →− 2
x →− 2
x →− 2
=9 ( −1) − 2 ( 7 ) =−23
Evaluate the Limits
(b)
lim 1 − 5 f ( x ) g ( x )  =
lim 1 − lim 5 f ( x ) g ( x )
x →− 2
x →− 2
x →− 2
= lim 1 − 5  lim f ( x )   lim g ( x ) 
 x →− 2
  x →− 2

x →− 2
=1 − 5 ( 8 )( −1) = 41
Propery 2
Property 3
Propery 7 & Evaluate
(c)
3
lim  f ( x ) 
 f ( x ) 
x →− 2 
=
lim
x →− 2 10 + h ( x )
lim (10 + h ( x ) )
3
Property 4
x →− 2
3
 lim f ( x ) 
 x →− 2

= 
lim 10 + lim h ( x )
x →− 2
x →− 2
[8]
=
10 + 7
512
17
3
=
4.
( z − 4 )( z + 2 ) + 2 − 3z =
lim
z →−1
z 2 + 5z + 4
Property 2 & 5
Propery 7 & Evaluate
( z + 1)( z − 6 ) = lim z − 6 = − 7
z 2 − 5z − 6
lim
= lim
z →−1 z 2 + 5 z + 4
z →−1 ( z + 1)( z + 4 )
z →−1 z + 4
3
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.
Math 2413
Homework Set 2
lim−
h
h →0
=lim−
−h
=lim − 1 =−1
h h →0 −
10 Points
because h < 0 in this case
h h →0
h
h
lim=
lim
1 1
= lim
=
h →0 + h
h →0 + h
h →0 −
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
8. In both of these limits the numerator is staying fixed at 4 and as x approaches 7 (from either side) we
can see that 7-x 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 7-x is positive since x < 7 and raising this to the fifth power will keep it positive. So, in
this case we have a fixed positive number in the numerator divided by something increasingly smaller
positive number and so the limit in this case will be ∞ .
In the second case 7-x is negative since x > 7 and raising this to the fifth power will keep it negative.
So, in this case we have a fixed n number in the numerator divided by an increasingly smaller negative
number and so the limit in this case will be −∞ .
Here are the official answers to this problem.
lim−
x →7
4
(7 − x)
5
= ∞
lim+
x →7
4
(7 − x)
5
= −∞