Basic Limit Exercises
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One-Sided Limits
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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
Two-Sided Limits
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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
Limits to Infinity
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Evaluate the following limits or state that the limit does not exist.
Solutions
Limits of Piecewise Functions
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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
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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)}
x
∈
[
1
,
9
]
{\displaystyle x\in \left[1,9\right]}
y
∈
[
2
,
10
3
]
{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}
c
∈
(
1
,
9
)
{\displaystyle c\in (1,9)}
f
(
c
)
=
3
{\displaystyle f(c)=3}
◼
{\displaystyle \blacksquare }
Notice
f
(
x
)
{\displaystyle f(x)}
x
∈
[
1
,
9
]
{\displaystyle x\in \left[1,9\right]}
y
∈
[
2
,
10
3
]
{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}
c
∈
(
1
,
9
)
{\displaystyle c\in (1,9)}
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)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
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}
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}}}
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
x
=
μ
{\displaystyle x=\mu }
f
(
9
4
π
)
≤
f
(
μ
)
≤
f
(
π
)
{\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}
μ
∈
[
π
,
9
4
π
]
{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}
◼
{\displaystyle \blacksquare }
Notice
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
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}
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}}}
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
x
=
μ
{\displaystyle x=\mu }
f
(
9
4
π
)
≤
f
(
μ
)
≤
f
(
π
)
{\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}
μ
∈
[
π
,
9
4
π
]
{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}
◼
{\displaystyle \blacksquare }
Solutions
External Links
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![{\displaystyle \lim _{x\to 2}{\Big [}4x^{2}-3x-1{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9090ef0539e55c8c495638167af1902f6c0371e)
![{\displaystyle \lim _{x\to 5}{\Big [}x^{2}{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddef2e7715fea4bcb0a1fdacbcc14b5049a50a46)
![{\displaystyle \lim _{x\to {\frac {\pi }{4}}}{\Big [}\cos ^{2}(x){\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2b628dde4a64c213ceea9a932bba29f50ebe270)
![{\displaystyle \lim _{x\to 1}{\Big [}5e^{x-1}-5{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fbd3c44c22819b3d1cfe0b427d665f4742f96bc)
![{\displaystyle \lim _{x\to 7^{-}}{\Big [}|x^{2}+x|-x{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/107d70cb67026562a0c3ed9db454d47178947b2b)
![{\displaystyle \lim _{x\to 1}\left[x^{2}+5x-{\frac {1}{2-x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ff82956e4812c90be92af6b92fe043d09fec3c)
![{\displaystyle \lim _{x\to -\infty }{\Big [}3x^{2}-2x+1{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a9cfb494972698a81cd5a92200191100d2baff1)
![{\displaystyle x\in \left[1,9\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c382323526a3226306c482a7e21229542942994b)
![{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67f7bb544ddc2a11b3554f7b3e3045c62bfd10f9)
![{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98277867d64ec64dd50b50a028ca57eaf0b9006e)
![{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39312b6220ae96b3cbd7da6667e8205f57963de8)
