2009
10.31

## The limit of a function f(x)

The limit of $\emph{f}(x)$ as $x$ approaches $a$ is the number $L$, written $\mathop {\lim }\limits_{x \to a} \emph{f}(x) = L$

provided the $\emph{f}(x)$ is arbitrarily close to $L$ for all $x$ sufficiently close to, but not equal to, $a$. A Limit is independent of the way (from the left or from the right) in which $x$ approaches $a$. Here are some properties of limits:

• $\text{If }\emph{f}(x) = c \text{ is a constant function, then}\mathop {\lim }\limits_{x \to a} \emph{f}(x) = \mathop {\lim }\limits_{x \to a} c = c$
• $\mathop {\lim }\limits_{x \to a} x^{n} = a^{n} \text{ for any positive integer } n$ $\text{If }\mathop {\lim }\limits_{x \to a} \emph{f}(x) = L_{1} \text{ and } \mathop {\lim }\limits_{x \to a} \emph{g}(x) = L_{2} \text{, where } L_{1} \text{ and } L_{2} \text{ are real numbers, then}$

• $\mathop {\lim }\limits_{x \to a} [ \emph{f}(x) \pm \emph{g}(x) ] = \mathop {\lim }\limits_{x \to a} \emph{f}(x) \pm \mathop {\lim }\limits_{x \to a} \emph{g}(x) = L_{1} \pm L_{2}$
• $\mathop {\lim }\limits_{x \to a} [ \emph{f}(x) \cdot \emph{g}(x) ] = \mathop {\lim }\limits_{x \to a} \emph{f}(x) \cdot \mathop {\lim }\limits_{x \to a} \emph{g}(x) = L_{1} \cdot L_{2}$
• $\mathop {\lim }\limits_{x \to a} [ c\emph{f}(x) ] = c \cdot \mathop {\lim }\limits_{x \to a} \emph{f}(x) = cL_{2} \text{, where } c \text{ is a constant}$
• $\text{If }\emph{f} \text{ is a polynomial function, then}\mathop {\lim }\limits_{x \to a} \emph{f}(x) = \mathop {\lim }\limits_{x \to a} \emph{f}(a)$
• $\mathop {\lim }\limits_{x \to a} \frac{ \emph{f}(x) }{ \emph{g}(x) } = \frac{ \mathop {\lim }\limits_{x \to a} \emph{f}(x) }{ \mathop {\lim }\limits_{x \to a} \emph{g}(x) } = \frac{ L_{1} }{ L_{2} } \text{, if } { L_{2} } \neq 0$
• $\mathop {\lim }\limits_{x \to a} \sqrt[n]{\emph{f}(x)} = \sqrt[n]{\mathop {\lim }\limits_{x \to a} \emph{f}(a)} = \sqrt[n]{L_{1}}$ *

* If $n$ is even, $L_{1}$ must be positive

Tip! Wolfram Alpha makes calculating limits really easy. Check this out.