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Level 3 Finding Limits Algebraically.notebook
1
May 08, 2014
Warm-Up:
1. Sketch a graph of the function and use the table to find the
lim 3x2 x⇒5
2. x⇒∞lim 3x2
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
finding limits algibraically
Ch.15.1 Level 3
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Target
Agenda
Purpose
Evaluation
TSWBAT: Find the limit of a function given Limit Laws.
Find the limit using Factoring, and Simplifying
Warm-Up
Lesson
BAT: Developing our Life Skill Get ready for calculus. Extend knowledge functions, make connections, apply knowledge to new situations
3-2-1
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
This is CRAZIER! By HAND!!!!
What is the lim 3x2 4x + 2?
What is the limit x4 + x2 2 ?44x
x⇒5
x⇒1
BIGGER
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Basic yet Necessary Rules
Finding the Limit Algebraically
lim C = C
lim x = a
lim xk = ak
lim √x = √akk
x⇒a
x⇒a
x⇒a
x⇒a
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
THE Limit Laws
lim [f(x) + g(x)] = M + Lx ⇒ a
Let c be a constant
lim f(x) = M and lim g(x) = Lx ⇒ a x ⇒ a
lim [f(x) g(x)] = M Lx ⇒ a
lim c f(x) = c Mx ⇒ a
lim [f(x) g(x)] = M Lx ⇒ a
lim x ⇒ a g(x)
f(x) M L=
1.
2.
3.
4.
5.
lim f(x)k = Mkx ⇒ a
6.
lim √f(x) = √Mx ⇒ a
7. k k
lim g(x)≠0
k > 0
a,k > 0
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Basically: Whatever you do to the function or functions, you can do directly to their limits
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
lim C = C
lim x = a
lim xk = ak
lim √x = √akk
x⇒a
x⇒a
x⇒a
x⇒a
Finding the Limit Algebraically
ex. lim 5 = x⇒16
ex. lim x = x⇒16
ex. lim (x4) = x⇒16
ex. lim x = x 16
Which rule did they use? Match them!
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Combining Limit Laws Example
lim 5(t + 3)8 x 4
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Summarizing
TIME!
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Finding the Limit Algebraically
Substitute a in for x, that's the limit!
lim f(x) = f(a) a Domain of fx a
ex. lim (3x2 4x + 2)x 6 **If this is true for all points in the
function, then it is continuous!
ex. lim 5(t + 3)8 x 4
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Canceling Common Factors
Use this tool when dividing by 0
Because: Law 5 Quotient and direct substitution cant be used to find the limits because we would be dividing by 0
ex. lim x 2 x2 4x 2
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Finding the Limit by SimplifyingUse when the numerator is not simplified or if you can cancel out the denominator completely
Because: Law 5 Quotient and direct substitution cant be used to find the limits because we would be dividing by 0
ex. lim (6+2x)2 36 2xx 0
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Finding the Limit by Finding:
Left and Right hand limits separately
Use when there is a piecewise function or absolute values in the function
Because:{ 3x if x < 0
x3 if 0 < x < 4
6x if x > 4
lim p(x)
p(x) =
x 0+
lim p(x)x 0
lim p(x)x 0
lim p(x)x 4+
lim p(x)x 4
lim p(x)x 3
lim p(x)x 4
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Finding the Limit by Finding:
Left and Right hand limits separately
Use when there is a piecewise function or absolute values in the function
Because:{ 4 if x=2
1/3x if x ≠ 2
lim p(x)
p(x) =
x 2+
lim p(x)x 2
lim p(x)x 2
p(2)
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Try it again!
Left and Right hand limits separately
Use when there is a piecewise function or absolute values in the function
Because:{ 1 if x < 2
3 if x=2
1 if x > 2
lim p(x)
p(x) =
x 2+
lim p(x)x 2
lim p(x)x 2
p(2)
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May 08, 2014
Evaluation:
1.Practice! pg.Worksheet
lim f(x) =x 4
(x2+5x + 6)(x+4) x2+6x+8
f(x) =
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Evaluation:
1.
Practice! pg.897 # 2734
lim f(x) =
x 4
|x+4| x216
f(x) =
2. lim f(x) =x 5
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Again! step it up!
Left and Right hand limits separately
Use when there is a piecewise function or absolute values in the function
Because:
ex. lim |x 3|x 3
x 0lim ( ) 1
2x 1 |2x|
+ex.
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
When the denominator has greater degree then...
lim 1 xx ∞
as x gets bigger, 1/x gets smaller 0, since ∞ isn't a number, we can get as close to 0 as we want
Special Infinity Rules
lim 1 xx ∞ k = 0 lim 1
xx ∞ k = 0
lim exx ∞ =0
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014
Special Infinity Rules Example
ex. f(x) = 4x3 x 3x3 + 2
ex. f(x) = 4x23 x 3x13 + 2x20
Level 3 Finding Limits Algebraically.notebook
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May 08, 2014