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Limits
If f(x) is defined for all x near a, except possibly at a itself, and if it can be ensured that f(x) is as close to L by taking x close enough to a, but not equal to a. Then f approaches the limit L as x approaches a:
\lim_{x \to a} f(x) = L
One-sided limits
If f(x) is defined on some interval (b,a) extending to the left of x=a, and if it can be ensured that f(x) is as close to L by taking x to the left of a and close enough to a, then $f(x) has left limit L at x=a and:
\lim_{x \uparrow a} f(x) = L.
If f(x) is defined on some interval (b,a) extending to the right of x=a and if it can be ensured that f(x) is as close to L by taking x to the right of a and close enough to a, then $f(x) has right limit L at x=a and:
\lim_{x \downarrow a} f(x) = L.
Limits at infinity
If f(x) is defined on an interval (a,\infty) and if it can be ensured that f(x) is as close to L by taking x large enough, then f(x) approaches the limit L as x approaches infinity and
\lim_{x \to \infty} f(x) = L
Limit rules
If \lim_{x \to a} f(x) = L, \lim_{x \to a} g(x) = M, and k is a constant then,
- Limit of a sum:
\lim_{x \to a}[f(x) + g(x)] = L + M. - Limit of a difference:
\lim_{x \to a}[f(x) - g(x)] = L - M. - Limit of a multiple:
\lim_{x \to a}k f(x) = k L. - Limit of a product:
\lim_{x \to a}f(x) g(x) = L M. - Limit of a quotient:
\lim_{x \to a}\frac{f(x)}{g(x)} = \frac{L}{M}, ifM \neq 0. - Limit of a power:
\lim_{x \to a}[f(x)]^\frac{m}{n} = L^{\frac{m}{n}}.
Formal definition of a limit
The limit \lim_{x \to a} f(x) = L means,
\forall \varepsilon > 0, \exists \delta \space \mathrm{,s.t.,} \space
\forall x \in \mathbb{R}, \space 0<|x-a|<\delta \implies |f(x) - L| < \varepsilon
For one-sided, infinite and limits at infinity there are similar formal definitions.
Example
Applying the formal definition of a limit for \lim_{x \to 4}\sqrt{2x + 1}
- Given
\varepsilon > 0 - Choose
\delta = \frac{\varepsilon}{2} - Suppose
0 < |x - 4| < \delta - Check
|\sqrt{2x + 1} - 3|
\begin{array}{ll}
|\sqrt{2x + 1} - 3| &= |\frac{(\sqrt{2x + 1} - 3)(\sqrt{2x + 1} + 3)}{\sqrt{2x + 1} + 3}|\
&= \frac{2|x - 4|}{\sqrt{2x + 1} + 3}\
&< 2|x-4|\
&< 2\delta = \varepsilon
\end{array}
Squeeze Theorem
Suppose that f(x) \leq g(x) \leq h(x) holds for all x in some open interval containing a, except possibly at x=a itself. Suppose also that
\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L.Then \lim_{x \to a} g(x) = L also. Similar statements hold for left and right limits.
Example
Applying squeeze theorem on \lim_{x \to 0} x^2 \cos(\frac{1}{x}).
\begin{array}{ll}
\forall x \neq 0\
-1 \leq \cos(\frac{1}{x}) \leq 1 \implies -x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2\
\mathrm{Since,} \space \lim_{x \to 0} x^2 = \lim_{x \to 0} -x^2 = 0\
\lim_{x \to 0} x^2 \cos(\frac{1}{x}) = 0
\end{array}