Solutions
Evaluate the following limits or state that the limit does not exist.
8.
lim
x
→
−
1
−
1
−
x
2
{\displaystyle \lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}}
The limit does not exist.
The limit does not exist.
Solutions
Evaluate the following limits or state that the limit does not exist.
12.
lim
x
→
4
1
x
−
4
{\displaystyle \lim _{x\to 4}{\frac {1}{x-4}}}
The limit does not exist.
The limit does not exist.
13.
lim
x
→
2
1
x
−
2
{\displaystyle \lim _{x\to 2}{\frac {1}{x-2}}}
The limit does not exist.
The limit does not exist.
20.
lim
x
→
0
|
x
|
x
{\displaystyle \lim _{x\to 0}{\frac {|x|}{x}}}
The limit does not exist.
The limit does not exist.
22.
lim
x
→
3
x
2
+
16
x
−
3
{\displaystyle \lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}}
The limit does not exist.
The limit does not exist.
25.
lim
x
→
3
x
+
3
x
2
−
9
{\displaystyle \lim _{x\to 3}{\frac {x+3}{x^{2}-9}}}
The limit does not exist.
The limit does not exist.
31.
lim
x
→
1
5
x
x
2
+
2
x
−
3
{\displaystyle \lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}}
The limit does not exist.
The limit does not exist.
32.
lim
x
→
−
1
ln
(
x
2
−
1
)
{\displaystyle \lim _{x\to -1}\ln({\sqrt {x^{2}-1}})}
The limit does not exist.
The limit does not exist.
33.
lim
x
→
1
arcsin
(
x
)
{\displaystyle \lim _{x\to 1}\arcsin(x)}
The limit does not exist.
The limit does not exist.
Solutions
Evaluate the following limits or state that the limit does not exist.
Solutions
Limits of Piecewise Functions [ edit | edit source ]
Evaluate the following limits or state that the limit does not exist.
48. Consider the function
f
(
x
)
=
{
(
x
−
2
)
2
if
x
<
2
x
−
3
if
x
≥
2
{\displaystyle f(x)={\begin{cases}(x-2)^{2}&{\text{if }}x<2\\x-3&{\text{if }}x\geq 2\end{cases}}}
c.
lim
x
→
2
f
(
x
)
{\displaystyle \lim _{x\to 2}f(x)}
49. Consider the function
g
(
x
)
=
{
−
2
x
+
1
if
x
≤
0
x
+
1
if
0
<
x
<
4
x
2
+
2
if
x
≥
4
{\displaystyle g(x)={\begin{cases}-2x+1&{\text{if }}x\leq 0\\x+1&{\text{if }}0<x<4\\x^{2}+2&{\text{if }}x\geq 4\end{cases}}}
50. Consider the function
h
(
x
)
=
{
2
x
−
3
if
x
<
2
8
if
x
=
2
−
x
+
3
if
x
>
2
{\displaystyle h(x)={\begin{cases}2x-3&{\text{if }}x<2\\8&{\text{if }}x=2\\-x+3&{\text{if }}x>2\end{cases}}}
Solutions
Intermediate Value Theorem [ edit | edit source ]
51. Use the intermediate value theorem to show that there exists a value
f
(
x
)
=
3
{\displaystyle f(x)=3}
for
f
(
x
)
=
x
+
1
x
{\displaystyle f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}}
from
1
≤
x
≤
9
{\displaystyle 1\leq x\leq 9}
. If you cannot use the intermediate value theorem to show this, explain why.
Notice
f
(
x
)
{\displaystyle f(x)}
is continuous from
x
∈
[
1
,
9
]
{\displaystyle x\in \left[1,9\right]}
. Ergo, the intermediate value theorem applies. For all
y
∈
[
2
,
10
3
]
{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}
, there exists a
c
∈
(
1
,
9
)
{\displaystyle c\in (1,9)}
so that
f
(
c
)
=
3
{\displaystyle f(c)=3}
.
◼
{\displaystyle \blacksquare }
Notice
f
(
x
)
{\displaystyle f(x)}
is continuous from
x
∈
[
1
,
9
]
{\displaystyle x\in \left[1,9\right]}
. Ergo, the intermediate value theorem applies. For all
y
∈
[
2
,
10
3
]
{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}
, there exists a
c
∈
(
1
,
9
)
{\displaystyle c\in (1,9)}
so that
f
(
c
)
=
3
{\displaystyle f(c)=3}
.
◼
{\displaystyle \blacksquare }
52. Use the intermediate value theorem to show that there exists an
x
=
μ
{\displaystyle x=\mu }
so that
f
(
μ
)
=
π
4
{\displaystyle f(\mu )={\frac {\pi }{4}}}
for
f
(
x
)
=
sin
(
x
)
−
cos
(
x
)
ln
x
{\displaystyle f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}}
from
π
≤
x
≤
9
4
π
{\displaystyle \pi \leq x\leq {\frac {9}{4}}\pi }
. If you cannot use the intermediate value theorem to show this, explain why.
Notice
f
(
x
)
{\displaystyle f(x)}
is continuous from
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
. Ergo, the intermediate value theorem applies.
f
(
π
)
=
1
ln
(
π
)
{\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}
f
(
9
4
π
)
=
0
{\displaystyle f\left({\frac {9}{4}}\pi \right)=0}
It is known the following is true:
1
≥
π
4
≥
0
{\displaystyle 1\geq {\frac {\pi }{4}}\geq 0}
. From there, we can directly argue the following:
1
ln
(
π
)
≥
1
⇒
1
ln
(
π
)
≥
1
≥
π
4
≥
0
⇒
1
ln
(
π
)
≥
π
4
≥
0
{\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}
By the intermediate value theorem, if
f
(
x
)
{\displaystyle f(x)}
is continuous from
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
, then there exists an
x
=
μ
{\displaystyle x=\mu }
so that
f
(
9
4
π
)
≤
f
(
μ
)
≤
f
(
π
)
{\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}
for
μ
∈
[
π
,
9
4
π
]
{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}
.
◼
{\displaystyle \blacksquare }
Notice
f
(
x
)
{\displaystyle f(x)}
is continuous from
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
. Ergo, the intermediate value theorem applies.
f
(
π
)
=
1
ln
(
π
)
{\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}
f
(
9
4
π
)
=
0
{\displaystyle f\left({\frac {9}{4}}\pi \right)=0}
It is known the following is true:
1
≥
π
4
≥
0
{\displaystyle 1\geq {\frac {\pi }{4}}\geq 0}
. From there, we can directly argue the following:
1
ln
(
π
)
≥
1
⇒
1
ln
(
π
)
≥
1
≥
π
4
≥
0
⇒
1
ln
(
π
)
≥
π
4
≥
0
{\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}
By the intermediate value theorem, if
f
(
x
)
{\displaystyle f(x)}
is continuous from
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
, then there exists an
x
=
μ
{\displaystyle x=\mu }
so that
f
(
9
4
π
)
≤
f
(
μ
)
≤
f
(
π
)
{\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}
for
μ
∈
[
π
,
9
4
π
]
{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}
.
◼
{\displaystyle \blacksquare }
Solutions