# Limit of a function

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The limit of a function is a fundamental concept in calculus and analysis concerning the behavior of the function near a particular value of its independent variable.

The limit of a function at $c$ is $L$ if for every $\varepsilon > 0$, there exists a $\delta > 0$ such that $0 < |x-c| < \delta$ implies $|f(x)-L| < \varepsilon$.

That is, if $f(x)$ is within any arbitrary distance of $L$ whenever $x$ is sufficiently close, but not necessarily equal to, $c$.

Note that $\varepsilon$ represents the greatest distance between the values of the function and its limit, while $\delta$ represents the distance from the values of the function's inputs to $c$.

The idea that the condition $f(c)=L$ is not required allows for the possibility that the values of function approaches $L$, but the actual value of $f(c)$ need not be $L$, or even defined.

If, by chance, we have $f(c)=L$, then we also say that $f$ is continuous at $c$.

## Sequential characterisation Edit

The limit of a function has an equivalent definition involving real sequences:

$l = \lim_{y \to x} f(y) \iff \forall (x_n)_{n \in \N} \subset D \setminus \{x\} : \lim_{n \to \infty} x_n = x, \lim_{n \to \infty} f(x_n) = l.$

### Proof Edit

1. Suppose $l = \lim_{y \to x} f(y)$, and $(x_n)_{n \in \N} \subset D \setminus \{x\}$ is an arbitrary sequence such that $\lim_{n \to \infty} x_n = x.$
Fix \begin{align}\epsilon > 0.\end{align} Now $\exists \delta > 0 : \forall y \in D, 0 < |y-x| < \delta \implies |f(y)-l|<\epsilon.$
Also, $\exists N \in \N : \forall n \in \N, n \ge N, |x_n - x|<\delta.$
As $x_n \ne x$, $n \ge N \implies 0 < |x_n - x| < \delta \implies |f(x_n) - l| < \epsilon.$
So, $\lim_{n \to \infty} f(x_n) = l.$
2. Now suppose $f(y)\nrightarrow l$ as $y \to x.$
Ie, $\exists \epsilon > 0 : \forall \delta > 0, \exists y \in D$ with \begin{align}0 < |y - x| < \delta\end{align} and $|f(x) - l| \ge \epsilon.$
So $\forall n \in \N,$ take $\delta = \frac{1}{n},$, so by the above
$\exists y_n \in D \setminus \{x\} : 0 < |y_n - x| < \frac{1}{n}$ and $|f(y_n) - l| \ge \epsilon.$
So we have found a sequence $(y_n)_{n \in \N} \subset D \setminus \{x\} : \lim_{n \to \infty} y_n = x, \lim_{n \to \infty} f(x_n) \ne l.$

The first paragraph shows the standard definition implies the sequential characterisation, and the second paragraph (via proof of the contrapositive) shows that the sequential characterisation implies the standard definition. Therefore the two are equivalent and either may be used as the definition.

## Properties of limitsEdit

Two-sided limits are only defined if the limit from the right it equal to the limit on the left. Examples of when this is not the case include peicewise functions or functions involving absolute values.

## Methods of limit evaluationEdit

### Numerical estimationEdit

The function $f(x) = \frac{x^2 - 1}{1-x}$ is undefined at x=1. However, we can find the value of the function at values very close to one.

 f(0.9) f(0.99) f(0.999) f(1.001) f(1.01) f(1.1) -1.9 -1.99 -1.999 -2.001 -2.01 -2.1

We can see from the table that the value is approaching 2, so we can infer that $\lim_{x \to 1} \frac{x^2 - 1}{1-x} = -2$

### Graphical estimationEdit

Limits can be estimated by graphing the function. For example, by looking at a graph of $\frac{\sin(x)}{x}$, one can clearly see that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$.

### Algebraic manipulationEdit

In some cases, limits can be found by algebraic manipulation. For example,

$\lim_{x \to 1} \frac{x^2 - 1}{1-x} = \lim_{x \to 1} -\frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} -(x+1) = -(1+1) = -2$

If the limit is to either positive or negative infinity, only the largest degree terms affect the final answer. For example,

$\lim_{x \to \infty} \frac{2x^2 + 3x + 2}{3x^2 + 4} = \lim_{x \to \infty} \frac{2x^2}{3x^2} = \lim_{x \to \infty} \frac{2}{3} = \frac{2}{3}$