Calculus and Analytical Geometry

Lecture # 5MTH 104

LimitsThe concept of limit is the fundamental building block on which all calculus concepts are based. We shall develop concept of limits both numerically and graphically and then techniques of evaluating limits will be discussed.

Tangent line

The line through P and Q is called a secant line for the circle at P. As Q moves along the curve towards P, then the secant line will rotate toward a limiting position. The line in this limiting position is called Tangent line.

Example Find an equation for the tangent line to the parabola at the point P(1,1).2xy

),( 2xxQ

secm

solution

Consider the secant line through P(1,1) and on the parabola that is distinct from P. The slope of this secant line is

112

sec

x

xm

1)1(

)1)(1(sec

xx

xxm

Q gets closer and closer to P means x gets closer and closer to 1

Pmxmy

P

at line tangent theof slope is where )1(1 line afor formula slopepoint the

: )1,1(at line tangent ofequation The

tan

tan

1)1(

)1)(1(sec

xx

xxm

.2 , line tangent of slope Thus tan m

The slope of secant line gets closer and closer to 2 as x gets closer and closer to 1

Hence the equation if tangent line is

12122221

)1(21

xyxy

xyxy

Limits---Numerically

1)( 2 xxxf

The most basic use of limits is to describe how a function behaves as the independent variable approaches a given variable. Consider

Examine the behavior of f (x) when the value of x gets closer and closer to 2?

x 1.5 1.9 1.95 1.99 1.995 1.999

f(x) 1.75 2.71 2.8525 2.9701 2.985025 2.997001

Approaching to 2 from the left side

Limits---Numerically

x 2.5 2.1 2.05 2.01 2.005 2.001

f(x) 4.75 3.31 3.1525 3.0301 3.015025 3.003001

Approaching to 2 from the right side

We write

3)1( 2

2lim

xxx

Limits---Graphically

3)1( 2

2lim

xxx

limits

Lxfax

)(lim

11 of valueabout the conjecture a make toevidence numerical Use lim

1x

xx

An informal view: If the values of f(x) can be made as close as we like to L by taking values of x sufficiently close to a ( but not equal to a), then we write

Example:

x 0.999 0.9999 0.99999 0 1.00001 1.0001 1.001

f(x) 1.9995 1.99995 1.999995 2.000005 2.00005 2.004988

Left side Right side

One sided limits

0 ,10 ,1

)( Consider xx

xx

xf

-1xx

and 1 limlim00

xxxx

One sided limits

Lxfax

)(lim

An informal view: If the values of f(x) can be made as close as we like to L by taking values of x sufficiently close to a ( but greater than a), then we write

And if the values of f(x) can be made as close as we like to L by taking values of x sufficiently close to a ( but less than a), then we write

Lxfax

)(lim

Relationship between one-sided and two sided limits

The two sided limit of a function f(x) exists at a if and only if both of the one sided limits exists at a and have the value; that is

)()(a

ifonly and if )( limlimlim-

x

xfa

LxfLxfx

ax

Infinite limits

Sometimes one sided or two sided limits fail to exist because the values of the function increases or decreases without bound.

xx

xx

1 and 1 limlim00

Infinite limits

3 and

32 Find limlim

33

x

xx

x

xx

Infinite limits

.)( write then we true,areboth Ifly.respective right, thefrom andleft thefrom approaches as bound without decreases )( that denote

)( and )( sexpression theSimilarly,

)( write then we true,areboth Ifly.respective right, thefrom andleft thefrom approaches x as bound without increases )( that denote

)( and )( sexpression The

lim

limlim

lim

limlim

xfaxxf

xfxf

xfaxf

xfxf

ax

axax

ax

axax

An informal view:

Vertical asymptotes

The line x = a is called a vertical asymptote of the curve y =f(x) if at least one of the following statements is true

limx a

f x

limx a

f x

limx a

f x

limx a

f x

limx a

f x

limx a

f x

Vertical asymptotes

Vertical asymptotes

Example515 )(

4313 )( of asymptote vertical theFind

3

2

5

xxb

xxxxa

Computing limitsWe will use limit laws to calculate limitsThe limit laws

even. is if 0 provided )()( )(

0 provided ,)(

)(

)()( )(

))())((( )(

)()( )(

)()( )(

)( )(

then,)( and )( that suppose and numbers, real be and let

11

22

1

21

21

21

21

limlim

limlim

lim

limlimlimlimlimlimlimlimlim

limlimlimlim

nLLx fxfvii

LLL

x g

x f

xgxfvi

LLx gx ff(x)g(x) v

LLx gx fg(x)f(x) iv

LLx gx fg(x)f(x) iii

axiikki

LxgLxfka

nn

ax

n

ax

ax

ax

ax

axaxax

axaxax

axaxax

axax

axax

The Squeezing Theorem

Lxfcxf

Lxhxgcxh

gccx

xhxfxghf, g

cx

cxcx

)( is, that , approaches aslimit thishas also then

)()( say , approaches aslimit same thehave and

If .at holdnot need esinequaliti that theexceptions possible with the,number thecontaining intervalopen somein allfor

)()()( satisfying functions be and let

lim

limlim

Example: show that .01sin2

0lim

xx

x

01sin

theoremsqueezingby Hence

.0)(,0)( )( and )(let

1sin

have we,11sin1 since

2

0x

0x0x

22

222

lim

lim lim

xx

xhxfxxhxxf

xx

xx

x

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits

Computing limits