21 KiB
21 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| List of limits | 2/3 | https://en.wikipedia.org/wiki/List_of_limits | reference | science, encyclopedia | 2026-05-05T08:15:28.347754+00:00 | kb-cron |
If
lim
x
→
c
f
(
x
)
=
lim
x
→
c
h
(
x
)
=
L
{\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}h(x)=L}
and
f
(
x
)
≤
g
(
x
)
≤
h
(
x
)
{\displaystyle f(x)\leq g(x)\leq h(x)}
for all x in an open interval that contains c, except possibly c itself,
lim
x
→
c
g
(
x
)
=
L
.
{\displaystyle \lim _{x\to c}g(x)=L.}
This is known as the squeeze theorem. This applies even in the cases that f(x) and g(x) take on different values at c, or are discontinuous at c.
== Polynomials and functions of the form xa ==
lim
x
→
c
a
=
a
{\displaystyle \lim _{x\to c}a=a}
=== Polynomials in x ===
lim
x
→
c
x
=
c
{\displaystyle \lim _{x\to c}x=c}
lim
x
→
c
(
a
x
+
b
)
=
a
c
+
b
{\displaystyle \lim _{x\to c}(ax+b)=ac+b}
lim
x
→
c
x
n
=
c
n
{\displaystyle \lim _{x\to c}x^{n}=c^{n}}
if n is a positive integer
lim
x
→
∞
x
/
a
=
{
∞
,
a
>
0
does not exist
,
a
=
0
−
∞
,
a
<
0
{\displaystyle \lim _{x\to \infty }x/a={\begin{cases}\infty ,&a>0\\{\text{does not exist}},&a=0\\-\infty ,&a<0\end{cases}}}
In general, if
p
(
x
)
{\displaystyle p(x)}
is a polynomial then, by the continuity of polynomials,
lim
x
→
c
p
(
x
)
=
p
(
c
)
{\displaystyle \lim _{x\to c}p(x)=p(c)}
This is also true for rational functions, as they are continuous on their domains.
=== Functions of the form xa ===
lim
x
→
c
x
a
=
c
a
.
{\displaystyle \lim _{x\to c}x^{a}=c^{a}.}
In particular,
lim
x
→
∞
x
a
=
{
∞
,
a
>
0
1
,
a
=
0
0
,
a
<
0
{\displaystyle \lim _{x\to \infty }x^{a}={\begin{cases}\infty ,&a>0\\1,&a=0\\0,&a<0\end{cases}}}
lim
x
→
c
x
1
/
a
=
c
1
/
a
{\displaystyle \lim _{x\to c}x^{1/a}=c^{1/a}}
. In particular,
lim
x
→
∞
x
1
/
a
=
lim
x
→
∞
x
a
=
∞
for any
a
>
0
{\displaystyle \lim _{x\to \infty }x^{1/a}=\lim _{x\to \infty }{\sqrt[{a}]{x}}=\infty {\text{ for any }}a>0}
lim
x
→
0
+
x
−
n
=
lim
x
→
0
+
1
x
n
=
+
∞
{\displaystyle \lim _{x\to 0^{+}}x^{-n}=\lim _{x\to 0^{+}}{\frac {1}{x^{n}}}=+\infty }
lim
x
→
0
−
x
−
n
=
lim
x
→
0
−
1
x
n
=
{
−
∞
,
if
n
is odd
+
∞
,
if
n
is even
{\displaystyle \lim _{x\to 0^{-}}x^{-n}=\lim _{x\to 0^{-}}{\frac {1}{x^{n}}}={\begin{cases}-\infty ,&{\text{if }}n{\text{ is odd}}\\+\infty ,&{\text{if }}n{\text{ is even}}\end{cases}}}
lim
x
→
∞
a
x
−
1
=
lim
x
→
∞
a
/
x
=
0
for any real
a
{\displaystyle \lim _{x\to \infty }ax^{-1}=\lim _{x\to \infty }a/x=0{\text{ for any real }}a}
== Exponential functions ==
=== Functions of the form ag(x) ===
lim
x
→
c
e
x
=
e
c
{\displaystyle \lim _{x\to c}e^{x}=e^{c}}
, due to the continuity of
e
x
{\displaystyle e^{x}}
lim
x
→
∞
a
x
=
{
∞
,
a
>
1
1
,
a
=
1
0
,
0
<
a
<
1
{\displaystyle \lim _{x\to \infty }a^{x}={\begin{cases}\infty ,&a>1\\1,&a=1\\0,&0<a<1\end{cases}}}
lim
x
→
∞
a
−
x
=
{
0
,
a
>
1
1
,
a
=
1
∞
,
0
<
a
<
1
{\displaystyle \lim _{x\to \infty }a^{-x}={\begin{cases}0,&a>1\\1,&a=1\\\infty ,&0<a<1\end{cases}}}
lim
x
→
∞
a
x
=
lim
x
→
∞
a
1
/
x
=
{
1
,
a
>
0
0
,
a
=
0
does not exist
,
a
<
0
{\displaystyle \lim _{x\to \infty }{\sqrt[{x}]{a}}=\lim _{x\to \infty }{a}^{1/x}={\begin{cases}1,&a>0\\0,&a=0\\{\text{does not exist}},&a<0\end{cases}}}
=== Functions of the form xg(x) ===
lim
x
→
∞
x
x
=
lim
x
→
∞
x
1
/
x
=
1
{\displaystyle \lim _{x\to \infty }{\sqrt[{x}]{x}}=\lim _{x\to \infty }{x}^{1/x}=1}
=== Functions of the form f(x)g(x) ===
lim
x
→
+
∞
(
x
x
+
k
)
x
=
e
−
k
{\displaystyle \lim _{x\to +\infty }\left({\frac {x}{x+k}}\right)^{x}=e^{-k}}
lim
x
→
0
(
1
+
x
)
1
x
=
e
{\displaystyle \lim _{x\to 0}\left(1+x\right)^{\frac {1}{x}}=e}
lim
x
→
0
(
1
+
k
x
)
m
x
=
e
m
k
{\displaystyle \lim _{x\to 0}\left(1+kx\right)^{\frac {m}{x}}=e^{mk}}
lim
x
→
+
∞
(
1
+
1
x
)
x
=
e
{\displaystyle \lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e}
lim
x
→
+
∞
(
1
−
1
x
)
x
=
1
e
{\displaystyle \lim _{x\to +\infty }\left(1-{\frac {1}{x}}\right)^{x}={\frac {1}{e}}}
lim
x
→
+
∞
(
1
+
k
x
)
m
x
=
e
m
k
{\displaystyle \lim _{x\to +\infty }\left(1+{\frac {k}{x}}\right)^{mx}=e^{mk}}
lim
x
→
0
(
1
+
a
(
e
−
x
−
1
)
)
−
1
x
=
e
a
{\displaystyle \lim _{x\to 0}\left(1+a\left({e^{-x}-1}\right)\right)^{-{\frac {1}{x}}}=e^{a}}
. This limit can be derived from this limit.
=== Sums, products and composites ===
lim
x
→
0
x
e
−
x
=
0
{\displaystyle \lim _{x\to 0}xe^{-x}=0}
lim
x
→
∞
x
e
−
x
=
0
{\displaystyle \lim _{x\to \infty }xe^{-x}=0}
lim
x
→
0
(
a
x
−
1
x
)
=
ln
a
,
{\displaystyle \lim _{x\to 0}\left({\frac {a^{x}-1}{x}}\right)=\ln {a},}
for all positive a.
lim
x
→
0
(
e
x
−
1
x
)
=
1
{\displaystyle \lim _{x\to 0}\left({\frac {e^{x}-1}{x}}\right)=1}
lim
x
→
0
(
e
a
x
−
1
x
)
=
a
{\displaystyle \lim _{x\to 0}\left({\frac {e^{ax}-1}{x}}\right)=a}
== Logarithmic functions ==
=== Natural logarithms ===
lim
x
→
c
ln
x
=
ln
c
{\displaystyle \lim _{x\to c}\ln {x}=\ln c}
, due to the continuity of
ln
x
{\displaystyle \ln {x}}
. In particular,
lim
x
→
0
+
log
x
=
−
∞
{\displaystyle \lim _{x\to 0^{+}}\log x=-\infty }
lim
x
→
∞
log
x
=
∞
{\displaystyle \lim _{x\to \infty }\log x=\infty }