18
LIMITS

Derivatives Lesson Oct 15

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Page 1: Derivatives Lesson  Oct 15

LIMITS

Page 2: Derivatives Lesson  Oct 15

We say that the limit of f(x), as x approaches a written

limx      a

f (x) = L

if we can take outputs f(x) as close to L as we wish by taking inputs x sufficiently close to a but not equal to a.

Page 3: Derivatives Lesson  Oct 15

Graph the function f(x) =  x2 ­ 9x ­ 3

Page 4: Derivatives Lesson  Oct 15

Now lets estimate  limx      3

f(x) 

On graphing calculator press 2nd [Tblset]

Page 5: Derivatives Lesson  Oct 15

Then press 2nd [TABLE]

Page 6: Derivatives Lesson  Oct 15

Graph f(x) =  sinxx

Page 7: Derivatives Lesson  Oct 15

estimate  limx      0

f(x)

Page 8: Derivatives Lesson  Oct 15
Page 9: Derivatives Lesson  Oct 15
Page 10: Derivatives Lesson  Oct 15

Graph f(x) = xx

Page 11: Derivatives Lesson  Oct 15
Page 12: Derivatives Lesson  Oct 15

limx      0+ x

x= 1 lim

x      0­ xx

= ­1

Right ­ hand limit:  limx      a+

f(x) = L means that f(x) can be 

made as close to L as we wish by taking x sufficiently close to a but greater than a

x      a­lim f(x) = L means that f(x) can be 

made as close to L as we wish by taking x sufficiently close to a but less than a

Left ­ hand limit:

Page 13: Derivatives Lesson  Oct 15

sinxx

limx      0+

sinxx

limx      0

sinxx

limx      0­

= = = 1

limx      0 x

xlimx      0+ x

x limx      0­ x

x= therefore,  does not exist

Page 14: Derivatives Lesson  Oct 15

Rules for calculation limits

1. Sum Rule: x      alim [ f(x) + g(x) ] =

x      alim f(x) +

x      alim g(x) = L + M

L ­ M

[ f(x) ­ g(x) ]2. Difference Rule:x      alim =

x      alim f(x) ­

x      alim g(x)

=

f(x) =x      alim g(x)3. Product Rule: x      a

limx      alim[ f(x)  g(x) ] = LM..

Page 15: Derivatives Lesson  Oct 15

4. Constant Multiple Rule:x      alim f(x)k = kL

5. Quotient Rule:x      alim =

f(x)g(x)

x      alim f(x)

x      alim g(x)

= LM

=if M      0

Page 16: Derivatives Lesson  Oct 15

Findx      2lim x2

3x + 4

x      2lim x2

3x + 4= x      2

lim x2

x      2lim (3x + 4)

= x      2lim x

x      2lim x

x      2lim

x      2limx3 4+

2

5

Page 17: Derivatives Lesson  Oct 15

Find x      2lim x2 ­ 5x + 6

x ­ 2

Page 18: Derivatives Lesson  Oct 15

Questions 1 ­ 19 odd only

Exercise 2.7