MATH 1910 Limits Worksheet Solutions
Find each of the following limits analytically and sketch a reasonable graph to support your
answer. If any limit does not exist, explain analytically why it does not exist. All answers must
be exact.
x2 + x − 6
=
x →2
x−2
1.
lim
lim
(x + 3) (x − 2)
(x − 2)
x →2
2.
3.
x2 − x + 6 8
Since the numerator
=
x →2
x−2
0
does not factor, we cannot cancel the
factor from the denominator.
No real limit exists.
⎛ x+2 +3⎞
⎜⎜
⎟⎟ =
x →7
⎝ x+2 +3⎠
x+2−9
lim
=
x → 7 (x − 7)( x + 2 + 3)
lim
x →7
x+2 −3
x−7
(x − 7)
(x − 7) ( x + 2 + 3)
=
3+x − 3 ⎛ 3+x + 3
⎜⎜
x →0
x
⎝ 3+x + 3
3+ x−3
lim
=
x → 0 x( 3 + x + 3 )
x →0
lim
x→4
lim
x→4
x
x ( 3 + x + 3)
x 2 − 16 ⎛ 2 + x ⎞
⎜
⎟
2 − x ⎜⎝ 2 + x ⎟⎠
=
⎞
⎟⎟ =
⎠
1
( 0, )
1
2 3
axes have been turned off
2 3
=
(x − 4) (x + 4)(2 + x)
(4 − x)
( 7, 16 )
1
6
lim
lim
5.
x →2
lim
lim
4.
= lim (x + 3) = 5
=
lim ( −1)(x + 4)(2 + x ) = −8(4) = −32
x→4
6.
x 2 + 2x − 8
0
=
=0
x →2
4
x 3 − 2x
This function is continuous at x = 2.
lim
Therefore, lim
x →2
x 2 + 2x − 8
= f(2) = 0
x 3 − 2x
( 2, 0 )
7.
1
1
− ⎛
2(2 + h) ⎞
2 − (2 + h)
lim 2 + h 2 ⎜
⎟ = lim
h→0
h
⎝ 2(2 + h) ⎠ h→0 2h(2 + h)
= lim
h→0
( 0, − 14 )
−1
1
=−
2(2 + h)
4
axes have been turned off
1
8.
1
−
9+h 3 ⎛ 3 9+h
⎜⎜
h
⎝ 3 9+h
lim
h→0
lim
h→0
⎞
⎟⎟ =
⎠
⎛ 3+ 9+h
⎜
h(3 + 9 + h ) ⎜⎝ 3 + 9 + h
3− 9+h
( 0, − 361 )
⎞
⎟⎟
⎠
axes have been turned off
= lim
h→0
= lim
h→0
9.
lim
x →0
9 − (9 + h)
h(3 + 9 + h )2
−1
(3 + 9 + h )
x
|x|
2
= lim
h→0
=−
−h
h(3 + 9 + h )2
1
36
There are 2 cases to consider!
Case 1:
If x > 0, then
x
= lim (1) = 1
lim
+
x →0 x
x →0 +
Case 2: If x < 0, then
x
lim
= lim ( −1) = −1
−
x x →0 −
−
x →0
x
x
x
Therefore, lim
DNE because lim
.
≠ lim
x →0 | x |
x →0 + | x |
x →0 − | x |
10.
|x+2|
There are 2 cases to consider!
x →−2 x + 2
Case 1: If x + 2 > 0, i.e. x > –2, then
x+2
lim
= lim (1) = 1
+
x →−2 x + 2
x →−2 +
Case 2: If x + 2 < 0, i.e. x < –2, then
−(x + 2)
lim
= lim ( −1) = −1
−
x+2
x →−2
x →−2 −
lim
Therefore, lim
x →−2
|x+2|
DNE because
x+2
lim
x →−2 +
|x+2|
|x+2|
.
≠ lim
−
x+2
x+2
x →−2
A removable discontinuity is a single point on a graph where the limit of the function exists but
the limit does not equal the value of the function at that point. This may be because the function
does not exist at that point, or it may be that the function has another value at that point. See
Cases 2 and 3 at the beginning of this handout for examples.
It is easy to redefine the function so that it is continuous at x = a; simply define f(a) = lim f(x) .
x →a
Here is a great website concerning limits: http://www.sasked.gov.sk.ca/docs/calc30/unit_c.htm
Example: g(x) =
x−2
x2 − 4
is clearly discontinuous at x = 2.
⎧ x−2
, if x ≠ 2
⎪⎪ 2
Define f(x) = ⎨ x − 4
. f(x) is continuous at x = 2 because f(2) = lim f(x) .
1
x→2
⎪
, if x = 2
⎪⎩
4
Redefine each of the following functions to remove the indicated discontinuity (if possible!)
x2 − 9
x−3
Based on the fact that
11. f(x) =
lim
(x − 3) (x + 3)
(x − 3)
x →3
at x = 3
= lim (x + 3) = 6 ,
x →3
⎧ x2 − 9
⎪
we redefine the function as r(x) = ⎨ x − 3 , if x ≠ 3
⎪⎩
6, if x = 3
⎧ x2 − 9
⎪
12. h(x) = ⎨ x − 3 , if x ≠ 3 at x = 3
⎪⎩
5, if x = 3
Based on the fact that
lim
(x − 3) (x + 3)
x →3
(x − 3)
= lim (x + 3) = 6 ,
x →3
⎧ x2 − 9
⎪
we redefine the function as r(x) = ⎨ x − 3 , if x ≠ 3
⎪⎩
6, if x = 3
13. g(x) =
x2 − 9
x 2 − 6x + 9
at x = 3
Based on the fact that lim
x →3
(x − 3) (x + 3)
(x − 3) (x − 3)
=
6
,
0
we cannot redefine the function to make it continuous at x = 3.
"We expect behavior at a point to be consistent with behavior near the point."
B. Long