Calculus I 2 Limits More

7/29/2019 Calculus I 2 Limits More

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SCC:Rickman More Notes on Limits. Page #1 of 4

Calculus IMore Notes on Limits.

TheSqueezeTheoremforRealLimits:The Squeeze Theorem for Real L imitsIf 1) a, b, c and L are real numbers,

2) a < c < b,

3) g(x) f (x) h(x) forall x in (a, c) (c, b)

,and 4)

x c x clim g(x) L lim h(x)

= =

then,x clim f (x) L

= .

The following graph demonstrates this theorem. In effect, since y = f(x) is bounded by y = g(x) and y = h(x), they funnel y = f(x) to Las x goes to c.

Example #1: If3

5 x f (x) 21 12x for all x in (0,5) , findx 2limf(x)

.

( )3 3x 2lim 5 x 5 (2) -3

= = ; ( )

x 2lim 21 12x 21 12(2) -3

= = . Therefore,

x 2lim f (x) -3

=

SpecialTrigonometricLimits:Example #2: Evaluate

x 0

sin(x)lim

x.

To find this limit we must look at 3 areas.

opp h

sin(x) h heighthyp 1

1Area base*height

2

1(1)*sin(x)

2

sin(x)

2

= = = =

=

=

=

The area and angle of the sector

above are proportional to thearea and angle of a full circle.

2

Area x

2(1)

Area x

2

xArea

2

=

=

=

opp h

tan(x) h heightadj 1

1Area base*height

2

1(1)*tan(x)

2

tan(x)

2

= = = =

=

=

=


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Now notice the relation between the 3 areas.

Thus,

sin(x) x tan(x)

2 2 2

sin(x) x tan(x)

sin(x) x and x tan(x)

At this point we need to work the 2 one-sided limits separately.

Case 1:x 0

sin(x)lim

x+. In this case we can restrict x to be between 0 and

2

since we are dealing with the right-hand limit and this limit

is only concerned with what is happening immediately to the right of 0.

andsin(x) x x tan(x)

|sin(x) sin(x)1 x|x cos(x)

| sin(x)cos(x)

x

sin(x)cos(x) 1

x

Case 2:x 0

sin(x)lim

x. Use only x between -

2

and 0. Thus,

andsin(x) x x tan(x)

|sin(x) sin(x)1 x

|x cos(x)

| sin(x)cos(x)x

sin(x)1 cos(x)

x

Now we just need thatx 0lim cos(x) cos(0) 1

= = , and

x 0lim1 1

= .

Thus, in either case the squeeze theorem would then conclude thatx 0

sin(x)lim 1

x+= , and

x 0

sin(x)lim 1

x= for both cases 1 and 2

respectively. Therefore,x 0

sin(x)lim 1

x= .

Example #3: Evaluatex 0

1 cos(x)lim

x

.

[ ] [ ]

[ ] [ ]

2

x 0 x 0 x 0

1 cos(x) 1 cos(x)1 cos(x) 1 cos (x)lim lim lim

x x 1 cos(x) x 1 cos(x)

+ = =

+ +

[ ] [ ] [ ]

2

x 0 x 0 x 0

sin (x) sin(x) sin(x) sin(0)lim lim lim (1) 0

x 1 cos(x) x 1 cos(x) 1 cos(0)

= = = = + + +

.


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Therefore,

The Special Trigonometric Limits

1)x 0

sin(x)lim 1

x=

and,

2)x 0

1 cos(x)lim 0

x

=

ProvingtheAreaofaCircle:

First we need the formula for the area of an isosceles triangle.

[ ]

2

opp h

sin(x) hyp r

r sin(x) h height

1Area base*height

2

1(r) * r sin( )

2

r sin( )

2

= =

= =

=

=

=

Now to approximate the area of a circle, we inscribe n isosceles triangles, of the same size, inside the circle as shown.

2

2

2

n

2n r sin

r sin( ) nApproximateArea n

2 2

using thefact thatallthe 's are thesame.

=

= =

( )2Thus,sincethis willalwaysunderapproximate thearea,

r sin xArea for all valuesof n.

x

Still we can improve the approximation by taking the limit as n .


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Therefore,

( )( )

( ) ( )

2

n

22

x 0 x 0

22

x 0 x 0

2

2

2n r sin

2 2nArea lim Now define x and thus, n

2 n x

2r sin x

2 r sin xxlim lim

2 2x

r sin x sin xlim r lim

x x

r (1)

r

= =

=

=

But this only proves that 2Area r .

So now repeat with circumscribed triangles.

In this case:

2

22 22 2

2 2

rcos

2 s

rs

cos2

rsin( )

r 2sin coscos r sin2 2s sin( ) r sin( )2 2

Areaof 1triangle2 2

2cos 2cos cos2 2 2

=

=

= = = = =

( )

( )

2 2 2

n n n

2

x 0 x

2 nr sin n r sin n r sin

2 2 nArea lim n lim lim Now define x and thus,n

2 n n xcos cos cos

2 2 n

r sin xxlim lim

cos x

= = = =

=

( )

( ) ( )

( )

( )

2 2 2

0 x 0 x 0

2

r sin x sin xr rlim lim (1)

x cos x cos x x cos 0

r

= =

Finally, we have that 2Area r , and 2Area r . Thus, 2Area r = .

I know that we used the formula for the area of a circle to getx 0

sin(x)lim 1

x= which causes these proofs together to be a circular

argument, butx 0

sin(x)lim 1

x= can be proved without using the area of a circle. Thus, ultimately the proof for the area of a circle is valid

I just cant yet show you a better proof ofx 0

sin(x)lim 1

x= using derivatives.

Calculus I 2 Limits More

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