Limits

$\lim_{x\to{a}}{f(x)}=L$ iff:

\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{x}\; (0<|x-a|<\delta\implies{|f(x)-L|<\epsilon})

Defining $\delta$ in terms of a given $\epsilon$ is enough to prove a limit.

Properties

Suppose $\lim f(x) = L, \lim g(x) =M$ .

\lim_{x\to{a}}{f(x)}=L \iff \Big( \lim_{x\to{a^-}}{f(x)}=L \land \lim_{x\to{a^+}}{f(x)}=L \Big)

$\lim_{x\to{a^{-}}}{f(x)}=L$ iff:

\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{x}\; (-\delta<x-a<0\implies{|f(x)-L|<\epsilon})

$\lim_{x\to{a^{+}}}{f(x)}=L$ iff:

\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{x}\; (0<x-a<\delta\implies{|f(x)-L|<\epsilon})

\lim_{x\to{a}}{f(x)}=L \iff \Big( \lim_{x\to{a}}{f(x)}=L^{+} \lor \lim_{x\to{a}}{f(x)}=L^{-} \Big)

$\lim_{x\to{a}}{f(x)}=L^{+}$ iff:

\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{x}\; (0<\lvert{x-a}\rvert<\delta\implies{0\le f(x)-L<\epsilon})

$\lim_{x\to{a}}{f(x)}=L^{-}$ iff:

\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{x}\; (0<\lvert{x-a}\rvert<\delta\implies{-\epsilon\lt f(x)-L\le 0})

$\lim_{x\to{\infty}}{f(x)}=L$ iff:

\forall{\epsilon\gt 0}\; \exists{N>0}\; \forall{x}\; (x\gt N\implies{|f(x)-L|<\epsilon})

$\lim_{x\to{-\infty}}{f(x)}=L$ iff:

\forall{\epsilon\gt 0}\; \exists{N>0}\; \forall{x}\; (x\lt-N\implies{|f(x)-L|<\epsilon})

$\lim_{x\to a}{f(x)}=\infty$ iff:

\forall{M\gt 0}\; \exists{\delta>0}\; \forall{x}\; (0<\lvert{x-a}\rvert<\delta\implies{f(x)\gt M})

$\lim_{x\to a}{f(x)}=-\infty$ iff:

\forall{M\gt 0}\; \exists{\delta>0}\; \forall{x}\; (0<\lvert{x-a}\rvert<\delta\implies{f(x)\lt-M})

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September 29, 2026 2 min read

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