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Limits and convergence

Calculus was invented almost simultaneously and independently by Isaac Newton and Gottfried Leibniz in the 1680s. Calculus is a tool which can be used to analyse and model change.

There are two basic types of calculus: differentiation and integration:

  • Differentiation
  • This is a technique for determining the slope of a tangent at a single point on a curve. It returns the rate of change of a function.

  • Integration
  • The opposite of differentiation. Integration returns the area under the graph of a curve.

Used in conjunction, these two forms of calculus constitute a powerful mathematical tool, making possible previously laborious, or even impossible, solutions.

Limits

Convergence of a series

The concept of limits is fundamental to calculus. Limits may refer to the sum of an infinite series:

$$2 + 1 + 1/2 + 1/4 + 1/8 + .... = 4$$

A limit refers to a convergence. A convergence is a value the expression approaches, but never quite reaches. Graphically, the convergence may result in an asymptote.

$${lim}↙{n→∞} ({1/2})^n = 0$$

In this limit, the value of $({1/2})^n$ approaches 0 as n gets very large. There is no rational value of n which gives a result that is exactly zero, but it is clear that as n gets larger, the result gets closer to zero.

Limits may be determined algebraically by splitting a complex numerator into parts, which makes it easier to see the values of the invidual limits:

e.g. ${lim}↙{x→∞} {4x - 3}/x = {lim}↙{x→∞} ({4x}/x - 3/x) $
$ = {lim}↙{x→∞} 4 - {lim}↙{x→∞} 3/x $
$ = 4 - 0 = 4$

Neat, isn't it?

Convergence

A function is convergent if it has a finite limit, and divergent if it does not (i.e. the limit approaches infinity).

When a function f(x) is taken to a limit, or arbitrarily close to a finite number n, then f(x) converges to n.

Take the expression $1/{x^2}$. Does this expression have a value as x approaches zero?

Limits can form asymptotes

As the denominator is made ever smaller, the value of the expression becomes ever larger. The limit is therefore a vertical asymptote, coincident with the y-axis, not a finite number, so the solution does not exist.

Similarly, if the limit were to positive and negative infinity ${lim}↙{x→±∞} 1/{x^2}$, then the x-axis becomes coincident with a horizontal asymptote.

As the denominator is made ever larger, the value of the expression becomes ever smaller. A second limit is therefore a horizontal asymptote, coincident with the x-axis, not a finite number, so the solution does not exist.

Convergence of a Series

The sum of a finite geometric series is: $S_n = {u_1(1-r^n)}/{1-r}$, where $r$ is the common ratio of two consecutive terms, and $n$ is the number of terms $u$.

For a geometric series, ${Σ}↙{n=0}↖{∞} = {lim}↙{n→∞} {u_1(1-r^n)}/{1-r}$.

When $-1 < r < 1$, ${lim}↙{n→∞} r^n = 0$, and the series converges to its sum, $S={u_1}/{1-r}$.

Definition of derivative

$$f'(x) = {lim}↙{h→0} {f(x + h) - f(x)}/{h}$$

$f'(x)$ means the first differential of the function f(x) and is the gradient of the tangent at that point. Another way of writing the differential of y with respect to x is Leibniz's notation: ${dy}/{dx}$

Mathematics question

A secant is a straight line between two points on a curve.

The tangent to a point on a curve is found by reducing the length of a secant between two points equidistant on either side of the point. At the limit where the length is zero, the secant is coincident with the tangent, so the gradient of the secant is the first differential of the curve at that point.

Write an expression for the gradient of a secant line for the function $f(x) = x^2 + 1$.

$f'(x) = {lim}↙{h→0} {f(x + h) - f(x)}/{h}$

$ = {lim}↙{h→0} {[(x + h)^2 + 1] - (x^2 + 1)}/h $

$ = {lim}↙{h→0}{h(2x + h)}/h = {lim}↙{h→0} 2x + h = 2x$

Continuity of functions

A function is said to continuous at $x=p$, if ${lim}↙{x→p}=f(p)$.

A function is continuous at $x=p$ if these conditions are met:

  1. $f$ is defined at $p$. $p$ must be an element of the domain of $f$.
  2. The limit of $f$ at $p$ exists.
  3. The limit of $f$ at $p$ is equal to the $f(x)$ at $p$.

An example of a discontinuous function is: $f(x) = \{ \table x-2, x<1; x +2, x>1$

As the function approaches 1 from the left ($x→1^{-}$), $f(x) → -1$, but from the right ($x→1^{+}$), $f(x) → 3$. The limit of $f$ at $p$ does not equal the value of $f(x)$ at $p$. Since $f(x)$ at $p$ has two possible values, the function is discontinuous.

L'Hôpital-Bernoulli theorem

This is a rather useful theorem for finding limits of continuous functions.

The de L'Hôpital-Bernoulli theorem:

Let $f(x)$ and $g(x)$ be two functions continuous in $[a;b]$ and derivable in $]a;b[$, and let $c$ be in $]a;b[$, and $g(x) ≠ 0$ for every $x$ in $]a;b[$ where $x ≠ c$. If ${lim}↙{x→c}f(x) = {lim}↙{x→c}g(x) = 0$ or $±∞$, then $${lim}↙{x→c}{f(x)}/{g(x)} = {lim}↙{x→c}{f'(x)}/{g'(x)}$$ if ${lim}↙{x→c}{f'(x)}/{g'(x)}$ exists.

Content © Andrew Bone. All rights reserved. Created : September 21, 2014

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