Limit Examples
1
1
θ
=
= = 1.
(1) lim θ csc θ = lim
sin θ
θ→0
θ→0 sin θ
1
lim
θ→0 θ
tan 2θ sin2 θ
2 sin 2θ sin θ sin θ
sin 2θ 1 sin θ sin θ
= lim
= 2 lim
=
3
θ→0
θ→0 2 θ cos 2θ θ
θ→0 2θ cos 2θ θ
θ
θ
θ
1
2 · 1 · · 1 · 1 = 2.
1
(2) lim
cos 3θ
1
= “ ” is undefined. (The left and right limits are −∞ and ∞,
θ→0
θ
0
respectively.)
(3) lim
cos θ
=?
θ→π/2 θ − π/2
Recall cos θ = − sin θ − π2 . Let q = θ − π2 .
− sin θ − π2
− sin q
Thus, ? = lim
= lim
= −1.
π
q→0
θ→π/2
θ− 2
q
(4) lim
θ sin θ
0·0
θ sec θ
= lim
=
= 0.
θ→0 cos θ
θ→0 csc θ
1
(5) lim
sin 3θ
3 θ sin 3θ
sin 3θ θ
= lim
= 3 lim
= 3 · 1 · 1 = 3.
θ→0 sin θ
θ→0 3 θ sin θ
θ→0 3θ sin θ
(6) lim
tan θ
1
1
1 1
1
tan θ
= lim
= lim
= · 2 = .
2
θ→0 2 sin θ cos θ
θ→0 sin 2θ
2 θ→0 cos θ
2 1
2
(7) lim
θ
θ
(8) lim
= lim
θ→0 θ + tan θ
θ→0 θ + tan θ
(9) lim cos arctan x = cos
x→∞
1
θ
1
θ
1
1
1
=
= .
tan
θ
θ→0
1+1
2
1+
θ
= lim
π
= 0.
2
tan x
is undefined. Why?
x→∞
x
(10) lim
(11) lim
1
x
x→∞
(12) lim+
x→π
(13)
+ sin θ
0 + sin θ
=
= tan θ. (Since θ does not depend on x.)
cos θ
cos θ
0 −∞
sin x csc x
x sin x + csc x
= lim+
+
= +
= −∞.
x→π
3x
3
3x
3
3π
lim
y→−π/2−
cot y − tan y
0−∞
=
= ∞.
y
−π/2
1
x+4
= 0.
x→∞ x2 − 16
(14) lim
x+4
1
=
lim
is undefined. (The left and right limits are −∞ and
x→4 x2 − 16
x→4 x − 4
∞, respectively.)
(15) lim
1
1
x+4
= lim
=− .
2
x→−4 x − 4
x→−4 x − 16
8
(16) lim
x+4
0+4
1
= 2
=− .
2
x→0 x − 16
0 − 16
4
(17) lim
4+0
4
4ex + 8 e−x
4 + 8e−x
4ex + 8
=
lim
=
lim
=
=
.
x→∞ 3ex + 7 e−x
x→∞ 3 + 7e−x
x→∞ 3ex + 7
3+0
3
(18) lim
4ex + 8
0+8
8
(19) lim
=
=
.
x→−∞ 3ex + 7
0+7
7
x2 + x + 1
π
x2 + x + 1
=
arctan
lim
arctan 1 = .
2
2
x→∞
x→∞ x + 3x + 4
x + 3x + 4
4
v
!2 r
u √
√
u
3+x
√
6x3 + x
6x3 + x
6x
t
lim
0 = 0.
(21) lim
=
lim
=
=
x→∞ x4 + 4x + 4
x→∞ x2 + 2
x→∞
x2 + 2
(20) lim arctan
r
x4 + 2x3 + 15x2 + 14x + 49 √
x2 + x + 7
= 1 = 1.
= lim
(22) lim √
x→∞
x→∞
x4 + 8
x4 + 8
1 + x1 + x72
x2 + x + 7
1
x2 + x + 7 x12
√
√
Or, lim
lim q
= √ = 1.
= lim
1 = x→∞
4
4
x→∞
x→∞
x +8
x + 8 x2
1
1 + 84
x
√
17x2 + 7 − 4x 17x2 + 7 + 4x
√
=
+ 7 − 4x = lim
(23) lim
x→∞
x→∞
1
17x2 + 7 + 4x
1
x + x7
17x2 + 7 − 16x2
x2 + 7
x
q
lim √
=
lim
=
= lim √
x→∞
x→∞
17x2 + 7 + 4x x→∞ 17x2 + 7 + 4x x1
17 + 7 + 4
√
17x2
√
x
∞+0
lim √
= ∞.
x→∞
17 + 0 + 4
√
16x2 + 7 − 4x 16x2 + 7 + 4x
√
(24) lim
+ 7 − 4x = lim
=
x→∞
x→∞
1
16x2 + 7 + 4x
1
7
16x2 + 7 − 16x2
7
x
x
q
lim √
= lim √
=
lim
x→∞
x→∞
16x2 + 7 + 4x x→∞ 16x2 + 7 + 4x x1
16 +
√
16x2
lim √
x→∞
√
√
2 + x − 4x
16x
16x2 + x + 4x
√
=
(25) lim 16x2 + x − 4x = lim
x→∞
x→∞
1
16x2 + x + 4x
1
16x2 + x − 16x2
1
x
x
lim √
= lim q
= lim √
1
x→∞
x→∞
16x2 + x + 4x x→∞ 16x2 + x + 4x x
16 +
lim √
+4
0
= 0.
16 + 0 + 4
√
√
x→∞
=
7
x
=
1
x
+4
1
1
= .
8
16 + 0 + 4
An Application. Graph y =
q
x 2+
3
x2
7x + 4
.
Solution. First we observe that there is a vertical asymptote at x = −4/7. Also
notice that the function is undefine at x = 0, but its behavior here is not clear. To
understand the behavior near x = 0 we will compute the limits as x goes to zero from
the left and right.
q
√
√
x 2 + x32
2x2 + 3
3
lim+
= lim+
=
.
x→0
x→0
7x + 4
7x + 4
4
q
√
√
x 2 + x32
− 2x2 + 3
− 3
lim
= lim−
=
.
x→0−
x→0
7x + 4
7x + 4
4
We got the negative sign in front of the squar root because x is negative. We
conclude that the function has a jump discontinuity at x = 0. Next we study the
behavior for large values of x.
q
q
3
x 2 + x32 1
x 2 + x2
x
= lim
lim
1 =
x→∞ 7x + 4
x→∞ 7x + 4
x
q
√
2 + x32
2
=
.
lim
x→∞ 7 + 4
7
x
q
q
x 2 + x32
x 2 + x32 1
x
lim
= lim
1
x→−∞ 7x + 4
x→−∞ 7x + 4
x
q
√
2 + x32
2
.
= lim
=
4
x→−∞ 7 +
7
x
Since we didn’t have to pull the x inside the square root symbol, the signs are the
same.
√
Lastly we will check to see if the function crosses its horizontal asymtote y = 27 .
Set,
q
√
x 2 + x32
2
=
7x + 4
7
and we if we can solve for x. Clearly x cannot be negative or zero so we assume x > 0.
√
√
2(7x + 4)
2x2 + 3 =
7
Squaring both sides gives
2(49x2 + 56x + 16)
.
2x2 + 3 =
49
Thus, 2x2 + 3 = 2x2 + 112x+32
. So, 147 = 112x + 32 and then x = 115
≈ 1.0267857.
49
112
We put all the information together to get the graph in Figure 1.
4
y 2
0
-4
-2
0
4
2
x
-2
-4
Figure 1. Plot of y =
q
x 2+
3
x2
7x + 4
.