Finding Limits Algebraically 5t2h5t

Finding Limits Algebraically - Classwork We are going to now determine limits without benefit of looking at a graph, that is lim f ! x" . x !a There are three steps to : 1) plug in a 2) FactorAcancel and go back to step 1 3) !, -!, or CDE 2x " 6 Example 1) find lim x 2 " 4 x + 1 Example 2) find lim x !"2 x !"2 x " 2 Gou can do this by plugging in. Gou can also do this by plugging in.

x2 " 2x " 8 x !"2 x2 " 4

x2 " 2x + 1 x !1 x3 " 1 0 Plug in and you get 0 - no good So attempt to factor and cancel

Example 3) find lim

Example 4) find lim

0 Plug in and you get 0 - no good

So attempt to factor and cancel

If steps 1 and 2 do not work (you have a zero in the denominator, your answer is one of the following:

!

-!

Does Not Exist (DNE)

To determine which, you must split your limit into two separate limits.: lim" f ! x " and lim f ! x " . Make a sign x !a

x !a +

chart by plugging in a number close to a on the left side and determining its sign. Gou will also plug in a number close to a on the right side and determine its sign. Each of these will be some form of !, either positive or negative. Rnly if they are the same will the limit be ! or -!. What this says is that in this case, lim" f ! x" = some form of # and lim f ! x" = some form of # x !a +

x !a

Gou need to check whether they are the same. 2x + 5 Example 5) find lim x !2 x " 2 9 Step 1) Plug in - 0 - no good Step 2) - Do factoringAcancel

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So your answer is !, -! or CDE

Stu Schwartz

Example 6)

find lim x !0

4 x2

4 Step 1) Plug in - 0 - no good

Step 2) - Do factoringAcancel

x2 + 2x " 3 x !"3 x 2 + 6 x + 9

So your answer is !, -! or CDE

2x " 4 x " 6 x 2 + 12 x " 8

Example 7) find lim

Example 8) find lim

%' x 2 " 4, x $ 1 find lim f ! x" Example 9) f ! x" = & x !1 '("2 x " 1, x % 1

% x ,x $ 2 '' x 2 " Example 10) f ! x" = & find lim f ! x" x !2 'x "3 '( x " 2 , x % 2

Example 11) lim x !0

x !2

3

x+2 " 2 x

Finally, we are interested also in problems of the type: lim f ! x" . Were are the rules: x !&#

Write f(x) as a fraction. 1) If the highest power of x appears in the denominator (bottom heavy), lim f ! x" = 0 x !&#

2) If the highest power of x appears in the numerator (top heavy), lim f ! x" = &# x !&# plug in very large or small numbers and determine the sign of the answer 3) If the highest power of x appears both in the numerator and denominator coefficient"of"numeratorYs"highest"power (powers equal), lim f ! x" = "coefficient"of"denominatorYs"highest"power x !&# 4 x 2 + 50 x !# x 3 " 85

Example 12) lim

Example 15)

4 x " 5x 2 + 3 x !"# 1 x lim

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4 x 3 " 5x 2 + 3x " 1 x !# 5x 3 " 7 x " 25

Example 14) lim

x 2 " 3x 2x + 1

Example 17) lim

Example 13) lim

Example 16) lim

x !#

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3x 3 " 23 x !# 4 x " 1

x !"#

x 2 " 3x 2x + 1

Stu Schwartz

Finding Limits Algebraically - Homework 1) lim 12

2) lim )

x !5

4)

lim 3x 2 " 4 x " 1 x !5

2x " 4 7) lim x !4 x " 1

5)

lim" 5x 3 " 7 x 2 + 2 x " 2

x !0

lim 4 x x !2

6) lim 3 y 4 " 6 y 3 " 2 y y !"1

x2 + 4x + 4 8) lim x !"2 x2

9) lim

t3 + 8 t !"2 t + 2

12) lim

x2 " 4x + 4 x2 + x " 6

x !1

2x " 2 x "1

x 2 " 16 x"4

11) lim

x2 + 6x + 5 x !"1 x 2 " 3x " 4

14) lim

x 3 + x 2 " 5x + 3 x 3 " 3x + 2

15) lim

x x "3

x x " 25

17) lim

y+6 y 2 " 36

18) lim

3" x x " 2x " 8

4 2 x " 2x + 1

20) lim

x x "5

21) lim

"x2 x2 " 6 x + 9

10) lim x !4

13) lim

16)

3)

x !0

lim x !5

19) lim x !1

2

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x !1

y !6

x !5

x !2

x !3

x !4

x !3

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2

Stu Schwartz

% x " 1, x $ 3 22) f ! x" = & find lim f ! x" x !3 '(2 x " 3, x % 3

%' x 3 " 1, x $ "1 find lim f ! x" 23) f ! x" = & x !"1 '(2 x , x % "1

%x "2 ,x $1 '' x " 1 24) f ! x" = & find lim f ! x" x !1 ' x '( x " 1 , x % 1

25) lim x !0

x+4 "2 x

% x 2 " 49 ' ,x * 7 27) f ! x" = & x " 7 ' 2 (k " 2, x = 7

%' x 2 " 2 x " 3, x * 2 26) Let f ! x" = & '(k " 3, x = 2 find k such that lim f ! x" = f !2"

find k such that lim f ! x" = f !7"

x !2

x !7

!

29) lim!"2 x + 11"

30)

2x " 3 31) lim x !# 4 x + 5

7 " 3x 3 32) lim 3 x !"# 2 x + 1

33) lim

2 x + 30 34) lim 12 x !"# 6 x " 5

35) lim

x !#

x !#

37) lim

x !#

x x2 + 4

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4x4 x !# 6 x 3 " 19

38)

lim

x !"#

x x2 + 4

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"

lim 3x 4 " 3x 3 + 5x 2 + 8 x " 3

28) lim 6

x !"#

x !#

2 5x " 3

4 x 2 " 3x " 2 " 5x 3 36) lim x !"# 9x2 + 9x + 7

39)

lim

x !"#

3x 2 + x x2 " 1

Stu Schwartz