l i m i t s !

LIMITS!Chapter 2 In The Calculus Book

WHAT IS A LIMIT?!

Limit gives us a language for describing how the outputs of a function behave as the inputs approach some particular value.

DON’T LOOK AT THE SUN!

A simple way to think of limits is the sun.If someone ask you to tell you the location of the sun, you cannot look directly at it, because your eyes will burn off (not really, but still).Same with limits. You can’t look directly at the problem or your eyes will burn off. :]You have to estimate a location that would best work.

STEPS FOR LIMITS!

1) Plug It In!

2) Factor, Then Plug It In!

3) Graph It!

TRY IT OUT!

1) lim (x3-2x2+1)x -2

2) lim x2-4x 3 x+2

3) lim x4+x3

x -1

ANSWER TIME!1) lim (x3-2x2+1)

x -2

((-2)3-2(-2)2+1)(-8-2(4)+1)-15

2) lim x2-4x 3 x+2

lim (x+2)(x-2)x 3 x+2(x-2)(3-2) 1

3) lim x4+x3

x -1

(-1)4+(-1)3

1-1 0

HINT! If You Can’t Figure Out The Problem Or Step 1 And 2 Don’t Work, Just Graph It! :D

FIND THE LIMIT!See If You Can Answer These FIVE Limits Questions!

1) lim g(x) x 3

2) lim g(x) x -2-

3) lim g(x) x 1

4) lim g(x) x 1+

5) lim g(x) x 2

HINT! The – After The Number Means It Comes From The Left. The + Comes From The Right!

ANSWER TIME!

1) lim g(x) = x 3

2) lim g(x) = x -2-

3) lim g(x) = x 1

4) lim g(x) = x 1+

5) lim g(x) = x 2

-2

0

DNE

2

0

CLICK ON THE SLIDE TO REVEAL THE

ANSWERS!

KNOW THESE THEROMS!

MEMORIZE THEM! lim sin□ =1

x 0 □ lim □ _ = 1

x 0 sin□ lim 1-cos□ =1

x 0 □

DETERMINE THE LIMIT!

Put those theroms in action with these 3 problems!

lim sin3x x 0 x

lim 1-cos7x x 0 x

lim x _ x 0 sin6x

ANSWER TIME!

lim sin3x [3] x 0 x [3]

lim 1-cos7x [7] x 0 x [7]

lim x [6] x 0 sin6x [6]

= 3(sin3x) =3 3x

= 7(1-cos7) =0 7x

= 6x =6 6(sin6x)

CLICK ON THE SLIDE TO REVEAL THE ANSWERS!

THE 3 STEPS TO TEST FOR CONTINUITY!

1) f(a) exists2) lim f(x) exists

x 0 3) f(a) = lim f(x)

x 0

WHICH ARE CONTINOUS?!

An Absolute Value Function?

A Step Function?

HINT! CONTINUOUS FUNTIONS ARE FUNCTIONS WHEN YOU DON’T HAVE TO PICK UP YOUR

PENCIL TO GRAPH IT!

YES! :]NO! A Step Function

Has Jump Discontinuity! :[

HOW ABOUT THESE?!

A Linear Function?

A Linear Function With A

Hole?

HINT! CONTINUOUS FUNTIONS ARE FUNCTIONS WHEN YOU DON’T HAVE TO PICK UP YOUR PENCIL

TO GRAPH IT!

YES! :]NO! A Linear Function With A Hole Has Point

Discontinuity :[

AND THIS?!

HINT! CONTINUOUS FUNTIONS ARE FUNCTIONS WHEN YOU DON’T HAVE TO PICK UP YOUR PENCIL

TO GRAPH IT!

A Rational Function?

NO! A Rational Function Has

Infinite Discontinuity! :[

TRY THE 3 STEPS OUT!

1) f(x)=x+2@ x=2?

2) f(x) x2-4 x-2

@ x=2?3) f(x) x2-4

x-2@ x=0?

HINT! REMEMBER THE THREE STEPS!TRY TO PROVE THAT THESE EXIST!You MUST Write All Three Steps Out!1) f(a) exists2) lim f(x) exists x0 3) f(a) = lim f(x)

x 0

ANSWER TIME!QUESTION UNO!f(x)=x+2@ x=2?1) f(x)=4(exists)

2) lim f(x) x0

lim f(x) =(x+2) x0

lim f(x)=((2)+2) x0

lim f(x) =4 (exists) x0

lim f(x)=((2)+2) x0

lim f(x) =4 (exists) x0

3) f(a) = lim f(x) x 0

4=4 (exists)

CONTINUOUS!

Always Write (exists) When it Exists!

ANSWER TIME!QUESTION DOS!f(x)=x2-4

x-2@ x=2?

1) f(x)=x2-4 x-2

f(x)=(2)2-4 (2)-2

f(x)=4-4 0

f(x)=DNE!

NOT CONTINUOUS!

ANSWER TIME!QUESTION TRES!f(x)=x2-4

x-2@ x=0?

1) f(x)=x2-4 x-2

f(x)=(0)2-4 (0)-2

f(x)=2 (exists)2) lim f(x) x0

lim f(x) = lim x2-4 x0 x0 x-2

lim f(x) = (x-2)(x+2) x0 x-2 lim f(x) = x+2 x0

lim f(x)=(0)+2 x0

lim f(x)=2 (exists) x0

3) f(a) = lim f(x) x 0

2=2 (exists)

CONTINUOUS!

How To Deal When x∞?

3 Short Cuts!1) Biggest Powered x On The

Denominator, ∞=02) Equal Powered x On Both The

Numerator And The Denominator, ∞=The Fraction Of The Two Coefficients From The Highest Powered x

3) Biggest Power x On The Numerator, ∞= ∞ or -∞

Try It Out!

1) lim 1 x∞ x

2) lim 3x2-5x+1 x∞ 4x2+3x+2

3) lim 3x4+2x+1 x∞ 2x3+x-2

HINT!1) Biggest Powered x On Bottom, ∞=02) Equal Powered x, ∞=The Fraction Of The Two Coefficients From The Highest Powered x3) Biggest Power x On Top,∞= ∞ or -∞

ANSWER TIME!1) lim 1 x∞ x lim = 0

x∞

2) lim 3x2-5x+1 [1/x2] x∞ 4x2+3x+2 [1/x2]lim 3-5/x+1/x2

x∞ 4+3/x+2/x2

lim = 3 x∞ 4

3) lim 3x4+2x+1 [1/x3] x∞ 2x3+x-2 [1/x3]lim 3x+2/x2+1/x3

x∞ 2+1/x2-2/x3

lim = ∞ x∞

Because x2=∞2, all the pink numbers would end up equaling 0. Therefore, cancelling them out.

Also Multiply By The Smallest Powered. In This Case, You Would Multiply By [1/x3], Not [1/x4]

If The ∞ Had A – In Front, The Answer Would Then Be Negative.If Either Coefficient Had A -, The Answer Would Also Be Negative.

CONGRATULATIONS! YOU ARE NOW READY

FOR THAT LIMITS TEST!