List of limits

${\displaystyle \lim _{x\to c}f(x)=L}$ if and only if ${\displaystyle \forall \varepsilon >0\ \exists \delta >0:0<|x-c|<\delta \implies |f(x)-L|<\varepsilon }$. This is the (ε, δ)-definition of limit.

The limit superior and limit inferior of a sequence are defined as ${\displaystyle \limsup _{n\to \infty }x_{n}=\lim _{n\to \infty }\left(\sup _{m\geq n}x_{m}\right)}$ and ${\displaystyle \liminf _{n\to \infty }x_{n}=\lim _{n\to \infty }\left(\inf _{m\geq n}x_{m}\right)}$.

A function, ${\displaystyle f(x)}$, is said to be continuous at a point, c, if

$${\displaystyle \lim _{x\to c}f(x)=f(c).}$$

If ${\displaystyle \lim _{x\to c}f(x)=L}$ then:

• ${\displaystyle \lim _{x\to c}\,[f(x)\pm a]=L\pm a}$
• ${\displaystyle \lim _{x\to c}\,af(x)=aL}$[1][2][3]
• ${\displaystyle \lim _{x\to c}{\frac {1}{f(x)}}={\frac {1}{L}}}$[4] if L is not equal to 0.
• ${\displaystyle \lim _{x\to c}\,f(x)^{n}=L^{n}}$ if n is a positive integer[1][2][3]
• ${\displaystyle \lim _{x\to c}\,f(x)^{1 \over n}=L^{1 \over n}}$ if n is a positive integer, and if n is even, then L > 0.[1][3]

In general, if g(x) is continuous at L and ${\displaystyle \lim _{x\to c}f(x)=L}$ then

• ${\displaystyle \lim _{x\to c}g\left(f(x)\right)=g(L)}$[1][2]

If ${\displaystyle \lim _{x\to c}f(x)=L_{1}}$ and ${\displaystyle \lim _{x\to c}g(x)=L_{2}}$ then:

• ${\displaystyle \lim _{x\to c}\,[f(x)\pm g(x)]=L_{1}\pm L_{2}}$[1][2][3]
• ${\displaystyle \lim _{x\to c}\,[f(x)g(x)]=L_{1}\cdot L_{2}}$[1][2][3]
• ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}={\frac {L_{1}}{L_{2}}}\qquad {\text{ if }}L_{2}\neq 0}$[1][2][3]

In these limits, the infinitesimal change ${\displaystyle h}$ is often denoted ${\displaystyle \Delta x}$ or ${\displaystyle \delta x}$. If ${\displaystyle f(x)}$is differentiable at ${\displaystyle x}$,

• ${\displaystyle \lim _{h\to 0}{f(x+h)-f(x) \over h}=f'(x)}$. This is the definition of the derivative. All differentiation rules can also be reframed as rules involving limits. For example, if g(x) is differentiable at x,
  • ${\displaystyle \lim _{h\to 0}{f\circ g(x+h)-f\circ g(x) \over h}=f'[g(x)]g'(x)}$. This is the chain rule.
  • ${\displaystyle \lim _{h\to 0}{f(x+h)g(x+h)-f(x)g(x) \over h}=f'(x)g(x)+f(x)g'(x)}$. This is the product rule.
• ${\displaystyle \lim _{h\to 0}\left({\frac {f(x+h)}{f(x)}}\right)^{1/h}=\exp \left({\frac {f'(x)}{f(x)}}\right)}$
• ${\displaystyle \lim _{h\to 0}{\left({f(e^{h}x) \over {f(x)}}\right)^{1/h}}=\exp \left({\frac {xf'(x)}{f(x)}}\right)}$

If ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ are differentiable on an open interval containing c, except possibly c itself, and ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty }$, L'Hôpital's rule can be used:

• ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$[2]

If ${\displaystyle f(x)\leq g(x)}$ for all x in an interval that contains c, except possibly c itself, and the limit of ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ both exist at c, then[5]

$${\displaystyle \lim _{x\to c}f(x)\leq \lim _{x\to c}g(x)}$$

If ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}h(x)=L}$ and ${\displaystyle f(x)\leq g(x)\leq h(x)}$for all x in an open interval that contains c, except possibly c itself,

$${\displaystyle \lim _{x\to c}g(x)=L.}$$

This is known as the squeeze theorem.[1][2] This applies even in the cases that f(x) and g(x) take on different values at c, or are discontinuous at c.

• ${\displaystyle \lim _{x\to c}x=c}$[1][2][3]
• ${\displaystyle \lim _{x\to c}(ax+b)=ac+b}$
• ${\displaystyle \lim _{x\to c}x^{n}=c^{n}}$ if n is a positive integer[5]
• ${\displaystyle \lim _{x\to \infty }x/a={\begin{cases}\infty ,&a>0\\{\text{does not exist}},&a=0\\-\infty ,&a<0\end{cases}}}$

In general, if ${\displaystyle p(x)}$is a polynomial then, by the continuity of polynomials,[5]

$${\displaystyle \lim _{x\to c}p(x)=p(c)}$$

This is also true for rational functions, as they are continuous on their domains.[5]

• ${\displaystyle \lim _{x\to c}x^{a}=c^{a}.}$[5] In particular,
  • ${\displaystyle \lim _{x\to \infty }x^{a}={\begin{cases}\infty ,&a>0\\1,&a=0\\0,&a<0\end{cases}}}$
• ${\displaystyle \lim _{x\to c}x^{1/a}=c^{1/a}}$.[5] In particular,
  • ${\displaystyle \lim _{x\to \infty }x^{1/a}=\lim _{x\to \infty }{\sqrt[{a}]{x}}=\infty {\text{ for any }}a>0}$[6]
• ${\displaystyle \lim _{x\to 0^{+}}x^{-n}=\lim {\frac {1}{x^{n}}}=+\infty }$
• ${\displaystyle \lim _{x\to 0^{-}}x^{-n}=\lim _{x\to 0^{-}}{\frac {1}{x^{n}}}={\begin{cases}-\infty ,&{\text{if }}n{\text{ is odd}}\\+\infty ,&{\text{if }}n{\text{ is even}}\end{cases}}}$
• ${\displaystyle \lim _{x\to \infty }ax^{-1}=\lim _{x\to \infty }a/x=0{\text{ for any real }}a}$

• ${\displaystyle \lim _{x\to c}e^{x}=e^{c}}$, due to the continuity of ${\displaystyle e^{x}}$
• ${\displaystyle \lim _{x\to \infty }a^{x}={\begin{cases}\infty ,&a>1\\1,&a=1\\0,&0<a<1\end{cases}}}$
• ${\displaystyle \lim _{x\to \infty }a^{-x}={\begin{cases}0,&a>1\\1,&a=1\\\infty ,&0<a<1\end{cases}}}$[6]
• ${\displaystyle \lim _{x\to \infty }{\sqrt[{x}]{a}}=\lim _{x\to \infty }{a}^{1/x}={\begin{cases}1,&a>0\\0,&a=0\\{\text{does not exist}},&a<0\end{cases}}}$

• ${\displaystyle \lim _{x\to \infty }{\sqrt[{x}]{x}}=\lim _{x\to \infty }{x}^{1/x}=1}$

• ${\displaystyle \lim _{x\to +\infty }\left({\frac {x}{x+k}}\right)^{x}=e^{-k}}$[2]
• ${\displaystyle \lim _{x\to 0}\left(1+x\right)^{\frac {1}{x}}=e}$[2]
• ${\displaystyle \lim _{x\to 0}\left(1+kx\right)^{\frac {m}{x}}=e^{mk}}$
• ${\displaystyle \lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e}$[7]
• ${\displaystyle \lim _{x\to +\infty }\left(1-{\frac {1}{x}}\right)^{x}={\frac {1}{e}}}$
• ${\displaystyle \lim _{x\to +\infty }\left(1+{\frac {k}{x}}\right)^{mx}=e^{mk}}$[6]
• ${\displaystyle \lim _{x\to 0}\left(1+a\left({e^{-x}-1}\right)\right)^{-{\frac {1}{x}}}=e^{a}}$. This limit can be derived from this limit.

• ${\displaystyle \lim _{x\to 0}xe^{-x}=0}$
• ${\displaystyle \lim _{x\to \infty }xe^{-x}=0}$
• ${\displaystyle \lim _{x\to 0}\left({\frac {a^{x}-1}{x}}\right)=\ln {a},}$ for all positive a.[4][7]
• ${\displaystyle \lim _{x\to 0}\left({\frac {e^{x}-1}{x}}\right)=1}$
• ${\displaystyle \lim _{x\to 0}\left({\frac {e^{ax}-1}{x}}\right)=a}$

• ${\displaystyle \lim _{x\to c}\ln {x}=\ln c}$, due to the continuity of ${\displaystyle \ln {x}}$. In particular,
  • ${\displaystyle \lim _{x\to 0^{+}}\log x=-\infty }$
  • ${\displaystyle \lim _{x\to \infty }\log x=\infty }$
• ${\displaystyle \lim _{x\to 1}{\frac {\ln(x)}{x-1}}=1}$
• ${\displaystyle \lim _{x\to 0}{\frac {\ln(x+1)}{x}}=1}$[7]
• ${\displaystyle \lim _{x\to 0}{\frac {-\ln \left(1+a\left({e^{-x}-1}\right)\right)}{x}}=a}$. This limit follows from L'Hôpital's rule.
• ${\displaystyle \lim _{x\to 0}x\ln x=0}$, hence ${\displaystyle \lim _{x\to 0}x^{x}=1}$
• ${\displaystyle \lim _{x\to \infty }{\frac {\ln x}{x}}=0}$[6]

For b > 1,

• ${\displaystyle \lim _{x\to 0^{+}}\log _{b}x=-\infty }$
• ${\displaystyle \lim _{x\to \infty }\log _{b}x=\infty }$

For b < 1,

• ${\displaystyle \lim _{x\to 0^{+}}\log _{b}x=\infty }$
• ${\displaystyle \lim _{x\to \infty }\log _{b}x=-\infty }$

Both cases can be generalized to:

• ${\displaystyle \lim _{x\to 0^{+}}\log _{b}x=-F(b)\infty }$
• ${\displaystyle \lim _{x\to \infty }\log _{b}x=F(b)\infty }$

where ${\displaystyle F(x)=2H(x-1)-1}$ and ${\displaystyle H(x)}$ is the Heaviside step function

Asymptotic equivalences, ${\displaystyle f(x)\sim g(x)}$, are true if ${\displaystyle \lim _{x\to \infty }{\frac {f(x)}{g(x)}}=1}$. Therefore, they can also be reframed as limits. Some notable asymptotic equivalences include

• ${\displaystyle \lim _{x\to \infty }{\frac {x/\ln x}{\pi (x)}}=1}$, due to the prime number theorem, ${\displaystyle \pi (x)\sim {\frac {x}{\ln x}}}$, where π(x) is the prime counting function.
• ${\displaystyle \lim _{n\to \infty }{\frac {{\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}{n!}}=1}$, due to Stirling's approximation, ${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}$.

The behaviour of functions described by Big O notation can also be described by limits. For example

• ${\displaystyle f(x)\in {\mathcal {O}}(g(x))}$ if ${\displaystyle \limsup _{x\to \infty }{\frac {|f(x)|}{g(x)}}<\infty }$