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by: Ren K.

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# MTH 132 Week 2 Lecture Notes MTH 132

Marketplace > Michigan State University > Mathematics > MTH 132 > MTH 132 Week 2 Lecture Notes
Ren K.
MSU
GPA 4.0

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These are all the notes from week 2 of MTH 132.
COURSE
Calculus 1
PROF.
Z. Zhou
TYPE
Class Notes
PAGES
12
WORDS
CONCEPTS
MTH, 132, MTH132, Calculus, Z.Zhou, Zhou, Math, Calc, engineers, Engineering
KARMA
25 ?

## Popular in Mathematics

This 12 page Class Notes was uploaded by Ren K. on Saturday September 10, 2016. The Class Notes belongs to MTH 132 at Michigan State University taught by Z. Zhou in Fall 2016. Since its upload, it has received 22 views. For similar materials see Calculus 1 in Mathematics at Michigan State University.

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Date Created: 09/10/16
MTH 132 ­ Lecture 2 ­ Limits    General definition  ● Let f(x൦) be a function near a. (a may not be in the domain)  ● We say that the limit of f(x൦) = L as x approaches a is denoted by limit f(x൦) = L  ● If f(x൦) is arbitrarily close to L by x sufficiently close to a (but not equal to) = the limit.       ○ We are interested in the behaviour of f(x൦) near a.  ○ Finding the limit has nothing to do with the value f(a).  Both Sides  ● For a limit to be defined ‘normally’ it has to have the same solution from both sides;  approaching from the positive side and approaching from the negative side.  ○ Otherwise it does not exist.   ○ Some can non exist because of oscillation.    ○ .    ● We can also define a one sided limit, where instead of approaching from both sides like  limit x approaches 1,  ○ we add a plus symbol to designate whether we’re approaching from the positive  side (the right)   ○ or a minus symbol to say we’re approaching from the negative side (the left).      ● If the limit exists from both sides, then the solution with the + and the solution with the ­  would be equivalent. Otherwise, if they’re different then the limit does not exist.  Finding limits!    MTH 132 ­ Lecture 3 ­ Approaching L    Review of previous concepts  ● Limit x→a f(x) = L  ● f(x) approaches L when x approaches a, but x =/ a.  ● f(x) doesn’t necessarily have to be defined.   ● Limit x → a+  ○ f(x) = L  ● Limit x → a­  ○ f(x) = L  ● If x→a+ and a­ are not equivalent then x→a limit DNE (Does not exist)    Limit Calculations  ● If  Limit x→a f(x)= L  ●       Limit x→a g(x) = M     ● Sum / Difference  ○ Limit x→a f(x) ± g(x) = L ± M   ● Constant Multiple  ○ Limit x→a [c*f(x)] = C * L  ● Product  ○ Limit x→a f(x)g(x) = L*M  ● Quotient  ○ Limit x→a f(x)/g(x) = L/M as long as M does not equal 0.  ● Power       ○ Limit x → a f(x)   = ( Limit x → a f(x))     ● Root   ○ Limit x → a √f(x) =√ Limit x → a f(x)   ○ When L doesn’t equal 0, you can have 3 different solutions  ■ If positive  ­ positive infinity  ■ If negative ­ negative infinity  ■ Or a DNE.     Composition equations  ● F ०g(x) = f(g(x))  ● Example:  ○  f(x) = sin x   2   ○ g(x) = x + 1   ○ f(g(x)) = sin (x + 1)   ● Example:  ○ Limit x → a g(x) = L  ○ Limit y→ L f(y) = M  ○ Limit x→a f(g(x)) not equal to M  ○ This may not be true if y is not equal to L  ● Example:  ○ g(x) = 0  ○ f(y) = 1 if y is not 0  ○          0 if y is 0.  ○ Limit y→0 f(y)=1      ● Limit x→0 f(g(x)) = 0  ● Limit x→ 0 g(x) =0 also because part of value is determined by gx  ● Example:  2   ● Limit x→a f (x)  =  L   ● Limit x→a [f(x)]  = L   ● N = 1,3    Squeeze theorem  ● If g(x)≤ f(x)  ≤ h(x) and Limit x → a g(x) = Limit x → a h(x) = L  ● Then f(x) = L        Oscillating limits with squeeze  ● f(x) = sin/ pi  ○  Limit x → 0   ○ f(x) = DNE  ● Limit x → 0   ● x  = 1/j  j ● f(xj) = sin j*pi = 0  ● Sin x j= 1 /( j + ½)  ● f(x )= sin(j+½)pi  j ● = − 1   ● ­g(x)≤ g(x) ≤ |g(x)| =  |x||sin pi/x|   ●                             *  =   |x| 1  = |x|  ○ Because sin always == 1      ● g(x) x sin(pi/x)  ● ­|x| ≤ |g(x)| ≤ |x|  ● Limit x ⟶ 0 g(x)  ○ The function is bounded between the positive absolute value of x and the  negative absolute value of x so its limit is ‘squeezed’ to equal zero.        Area of an Angle  ● Lim ө ⟶0 (sin  ө)/ ө  ● Whole of Unit circle = 2pi  ● Area = pi; comes from area circle formula.  ● Area of an angle  ө/2      ● Limit ө⟶0  ө/ sinө = 1  ● Limit  ө⟶ sinө/ ө = limit  ө⟶0 1/ ө/sin ө = 1/1 = 1  ● Limit x ⟶0 sin(3x)/x = limit x ⟶0  ⟶sin(3x)/3x * 3 = 3  MTH 132 ­ Lecture 4 ­    ● We’ve studied that the limit as x approaches a for f(x) = L   ○ ⇔ f(x) can get arbitrarily close to L when x ⇢a but x≠a.  ● For any  ε > 0, we can find ▯>0 such that |f(x) − L|< ε for all 0 <|x−a|<▯   ○   ● Example:  ○ Limit x → 2 (3x+2) = 8  ○ |f(x) ­ 8| ＜ ε  ○ ⇔|3x + 2 ­ 8| <ε  ○ ⇔|3x ­ 6| < ε  ○ ⇔3|x­2|<ε  ○ ⇔|x­2|<ε/3  ○ ⇔­ε/3 < |x­2 < ε/3  ● Example:  ○ Limit x ➝ for  x√= 3  ○ Proof: For any ε>0 we want | x­3|√<ε  ○ ­ε< | x√3| <ε  ○ 3­ε|  x√| < 3+ε  2 2 ○ (3­ε)  < x < (3+ε)    ○ 9 ­ 6ε + ε  < x < ε  + 6ε + 9  2 2 ○ ­6ε + ε  < x ­ 9 < ε  + 6ε        2 ● Choose ▯ = min{ ε + 6ε, 6ε − ε , ε(6ε−ε)}  2

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