Limits and the derivative function

Limits and the derivative function


The Velocity Problem

A particle is moving in a straight line.

t is the time that has passed from the start of motion (whichcorresponds to t = 0)

s(t) is the distance from the particle to the initial position at thetime t (s(t) is called the displacement function)

t = 0

s(t)

initialposition

position attime t

How to find the velocity of the particle at a given moment of time?



We can define the average velocity of the particle on a particulartime interval.Fix a moment of time t and consider the time interval [t, t +∆t].

0s(t)

t

s( )

t + ∆t

t + ∆t

distance traveledover the time interval[t, t + ∆t]

The distance traveled over the period [t, t +∆t] is s(t +∆t)− s(t).Hence, the average velocity of the particle on the fixed interval is

s(t +∆t)− s(t)

∆t



Now, the (instantaneous) velocity v(t) of the particle at the time t

is equal to the average velocity on [t, t +∆t] as the length of thetime interval ∆t decreases to 0.

s(t +∆t)− s(t)

∆t−→ v(t) as ∆t −→ 0

In this case we write

v(t) = lim∆t→0

s(t +∆t)− s(t)

∆t


The Tangent Line Problem

Given the graph of a function f (x), we would like to find thetangent line at the point P(x0, f (x0)).




0

y = f(x)

x

y

x0

f( )x0P

tangentline




0

y = f(x)

x

y

x0

f( )x0P

tangentline




0

y = f(x)

x

y

x0

f( )x0P

tangentline



The tangent line only “touches” the graph of f (x) at P but doesnot intersect it anywhere close to P .

It can be obtained as a limiting position of secant lines as follows





0

y = f(x)

x

y

P





















0

y = f(x)

x

y

P



Each secant line passes through P and some point Q(x1, f (x1)) onthe graph of f (x).

0

y = f(x)

x

y

P

Q

x0 x1

f( )x0

f( )x1

∆x

∆fQ

Q

Its slope is a ratio ∆fQ∆xQ

As Q approaches P , the ratio ∆fQ∆xQ

approaches the slope of the

tangent line at P .



In other words

∆fQ

∆xQ−→ slope of the tangent line as Q −→ P ,

recall that ∆fQ = f (x1)− f (x0), ∆xQ = x1 − x0, so

f (x1)− f (x0)

x1 − x0−→ slope of the tangent line as x1 −→ x0,

or,

slope of the tangent line = limx1→x0

f (x1)− f (x0)

x1 − x0


The Area Problem

Given a circle C of radius r , how to compute its area?

The idea is to approximate the circle by figures whose areas are easyto compute.

The easiest way is to use regular polygons Pn with n sides (n-gons)inscribed in the circle C .

The area of each Pn is easy to express in terms of the radius of thecircle

As n → ∞ the area of the n-gon Pn approaches the area of thecircle.


The Area Problem

Area(Pn) −→ Area(C ) as n −→ ∞,


The Area Problem

C



The Area Problem

P4

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The Area Problem

P5

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The Area Problem

P6


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The Area Problem

P8

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The Area Problem

P12

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Informal definition of limit

Definition

The limit of a function f (x), as x approaches a ∈ R, is L ∈ R, and we

write

limx→a

f (x) = L

if the values of f (x) can be made arbitrarily close to L by choosing the

values of x close enough to a.

In other words, the statement

limx→a

f (x) = L

means that we can make the distance between f (x) and L arbitrarilysmall by making the distance between x and a small enough but notequal to 0.That is, x approaching a makes the corresponding value f (x) approach L.


Formal definition of limit

1 First of all, recall that the distance between f (x) and L is |f (x)− L|,and the distance between x and a is |x − a|.

2 To make |f (x)− L| “arbitrarily small” means that |f (x)− L| can bemade smaller than any given positive real number, say ε.

3 Once ε > 0 is given, we have to choose a bound δ on the distance|x − a| small enough to force |f (x)− L| < ε.

4 Hence the precise definition of limit

Definition

We write

limx→a

f (x) = L

if for any ε > 0 there exists δ > 0 such that

|x − a| < δ implies |f (x)− L| < ε.


Examples

Example

limx→1 1 = 1

Example

limx→1x2−1

x−1= 2

Example

limx→01

x2= undefined


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Limits and the derivative function