# Difference between revisions of "Limit"

A limit is the behavior of a function as its inputs approach arbitrarily close to a given value.

Limits are written in the following form:

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

The expression above states that when $x\,$ approaches arbitrarily close to $a\,$, the function $f(x)\,$ becomes arbitrarily close to the value $L\,$, which is called the limit.

## Informal Definition

We can examine the limit of a simple continuous function, $f(x)=x^2 \,$.

We want to examine the limit of x= 0 for this function. Since this graph is a simple unbroken line, we realize that

$\lim_{x \to 0}f(x) = 0$

Indeed for this function, $\lim_{x \to a}f(x) = f(a) \,$. But this is a special case, in the majority of limits cannot be solved in this manner.

For a very different example; given

$f(x)=\left\{\begin{matrix} {x}^2, & \mbox{if }x\ne 0 \\ \\ 1, & \mbox{if }x=0. \end{matrix}\right.$ (as pictured below)

The limit of $f(x)\,$ because x approaches 0 is 0 (just as in $f(x)\,$), but $\lim_{x\to 0}f(x)\neq f(0)$; $f(x)$ is not continuous at $x = 0$ (as shown on the right).

In other cases a limit can fail to exist, as approaching the limit from different sides produces conflicting values.

Here we look at one such case:

$f(x)=\left\{\begin{matrix} {x}^2, & \mbox{if }x\ne 0 \\ \\ 1, & \mbox{if }x=0. \end{matrix}\right.$ (as pictured above)

## Rigorous Definition of Limit

This definition is more appropriate for 2nd year calculus students and higher.

The original statement $\lim_{x \to a}f(x) = L$ now means that given any $\varepsilon > 0$, a $\delta > 0 \,$ exists such that if $0 < |x-a| < \delta \,$, then $|f(x)-L|< \varepsilon$.

What this means to us is that for every region of positive radius \varepsilon at a given point, there is a

This definition is more useful in higher mathematics work because it eliminates the phrases "arbitrarily small" and "sufficiently small" from our intuitive definition. These phrases cannot be readily defined in mathematics language.

We use the definition as follows to prove for the function $f(x) = 3x-1 \,$, $\lim_{x \to 2}f(x) = 5$

Given any $\delta\,$, we choose a $\varepsilon\,$ such that $\delta \leq \varepsilon/3$

If $|x-2|<\delta\,$, we can derive $|f(x)-5|<3\delta\,$.

$3|x-2|<3\delta\,$

$|3x-6|<3\delta\,$

$|f(x)-5|<3\delta\,$

$3\delta \leq 3*\varepsilon/3 = \varepsilon\,$

Thus $\lim_{x \to 2}f(x) = 5$

## Properties of Limits

The limit of a sum of two functions is equal to the sum of the limits of the functions.

$\lim_{x \to a} (f(x)+g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$

The limit of a difference between two functions is equal to the difference between the limits of the functions.

$\lim_{x \to a} (f(x)-g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$

The limit of a product of two functions is equal to the product of the limits of the functions.

$\lim_{x \to a} (f(x)*g(x)) = \lim_{x \to a} f(x) * \lim_{x \to a} g(x)$

The limit of a quotient of two functions is equal to the quotient of the limits of the functions (assuming a non-zero denominator).

$\lim_{x \to a} (\frac{f(x)}{g(x)}) = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$

## Ideas for the Future

- an interactive diagram in which changing the size of epsilon shows a corresponding delta, or something.