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Learning Calculus II: Limits’ laws for functions and related theorems

2. Limits’ properties and related theorems

We can see that it’s really hard to find the limit of a complex function by using formal definition directly. Hence, here is some limits’ properties and theorems that can be useful in finding limit. Firstly, let’s rewrite definition of limits:

Definition 11 Let {f} be a function defined on an open interval {D} that contains {a}, except possibly at {a}. We say the limit of {f(x)} as {x} approaches {a} is {\ell}, and we write

\displaystyle \lim_{x \rightarrow a}f(x)=\ell

if for every number { \varepsilon>0} there is a number {\delta>0} such that:

if for all {0<|x-a|< \delta} and {x \in D} then {|f(x)-\ell| < \varepsilon}.

2.1. Limit laws for functions

Theorem 12 (Limit laws for functions) Let {f,g} be functions defined on interval {D} that contains {a}. Suppose {f,g} have limit {L,G} at {x=a}. Then

  • (a) {f \pm g} has limit {L \pm G} at {x=a}. We can write

    \displaystyle \lim_{x \rightarrow a}(f \pm g)(x)=\lim_{x \rightarrow a}f(x)+ \lim_{x \rightarrow a}g(x).

  • (b) {\max(f,g)} has limit {\max (L,G)} at {x=a}. Similarly for {\min (f,g)}. We can write

    \displaystyle \lim_{x \rightarrow a} \max (f,g)(x) = \max \left( \lim_{x \rightarrow a} f(x), \lim_{x \rightarrow a} g(x) \right).

  • (c) If {c} is a real number, then {cf} has a limit {cL} at {x=a}. We can write

    \displaystyle \lim_{x \rightarrow a} cf= c \lim_{x \rightarrow a}f(x).

  • (d) {fg} has limit {LG} at {x=a}. We can write

    \displaystyle \lim_{x \rightarrow a} (fg)(x)= \lim_{x \rightarrow a}f(x) \lim_{x \rightarrow a}g(x).

Proof: (a) It suffices to prove {\lim_{x \rightarrow a} (f+g)(x)=L+G}, i.e. according to definition 11 for any {\varepsilon>0} there is {\delta>0} so {|g(x)+f(x)-L-G|<\varepsilon} for all {0<|x-a|<\delta}. We need to use given condition that {\lim_{x \rightarrow a} f(x)=L} and {\lim_{x \rightarrow a}g(x)=G}, i.e. for any {\varepsilon_1, \varepsilon_2} there exists {\delta_1,\delta_2>0} so that {|f(x)-L|<\varepsilon_1} for all {0<|x-a|<\delta_1} and {|g(x)-G| \varepsilon_2} for all {0<|x-a|<\delta_2}. Note that

\displaystyle |f(x)+g(x)-L-G| \le |f(x)-L|+|g(x)-G| <\varepsilon_1+\varepsilon_2.

Hence, we can pick {\varepsilon_1,\varepsilon_2} so that {\varepsilon=\varepsilon_1+\varepsilon_2}. This will follow {|f(x)+g(x)-L-G| <\varepsilon} for all {0<|x-a|< \min (\varepsilon_1,\varepsilon_2)=\delta}. Thus, (a) is proven according to definition 11.



Learning Calculus I: Definitions of limits

I am learning calculus now I want to type up everything I now about this subject. All solutions written below, if no source is mentioned, are my solutions so if someone find an error in the solutions, please tell me. By the way, thanks to Luca Trevisan’s LaTeX to WordPress, I now don’t have to write LaTeX in WordPress format anymore, which is very convenient.

1. Definition of limits

1.1. Definition of limits of function

For me, the precise definition of limits is a bit hard to understand so I decided to reconstruct the whole thing by writing some of my guesses on why did mathematicians write the definition in this way. By the way, French mathematician, Augustin-Louis Cauchy (1789-1857) is the one who introduced {\delta, \varepsilon} into definition of limits.

Limit, the notation {\lim_{x \rightarrow a}f(x)=\ell} is understood vaguely as: as {x} gets closer to {a}, {f(x)} gets closer to a number {\ell}. What does it mean by ‘get closer’? It is easier to understand if we imagine {a} lies on a number line and suppose we pick an arbitrary point {x} on the line. If we apply the term ‘{x} gets closer to {a}‘, we can understand it as distance from {x} to {a} on the number line decreases. Its distance is {|x-a|} so saying {x} gets closer to {a} is the same as saying {|x-a|} gets closer to {0}. Note that for any distance {|x-a|}, there are two possible {x} on the number line that can satisfy that distance. What we care about is the distance from {x} to {a} not the position of {x}, so using ‘{|x-a|} gets closer to {0}‘ is better than ‘{x} gets closer to {a}‘. Similarly to {|f(x)-\ell|} and we get a new definition:

Definition 1 The symbol {\lim_{x \rightarrow a}f(x)=\ell} means that as {|x-a|} gets closer to {0} then {|f(x)-\ell|} gets closer to {0}.

Note that we can get {|x-a|} as close to {0} as we want by choosing appropriate {x}, but we can’t guarantee the same thing for {|f(x)-\ell|} since {f(x)} depends on {x}. Thus, firsly, in order for {|f(x)-\ell|} to be able to approach {0} we need {|f(x)-\ell|} to be as small as we want. It is equivalent to saying:

For any {\varepsilon>0} there exists some {x} so {\varepsilon>|f(x)-\ell|}.

Secondly, we need the definition to say {|f(x)-\ell|} approaches {0} as {|x-a|} approaches {0}. This means for small enough {\delta}, {|f(x)-\ell|} must be bounded by some {\varepsilon} for all {|x-a|<\delta}. For better understanding more about this argument, let’s assume the contrary, then as {\delta} gets closer to {0}, {|f(x)-\ell|} still does not have a bound for all {|x-a|<\delta}. Say we find at {|x-a|=\delta_1<\delta} has {|f(x)-\ell |=M_1}. We consider all {x} in {|x-a|<\delta_1} and follow that there must be an {x} so {|x-a|=\delta_2<\delta_1} and {|f(x)-\ell|=M_2>M_1} according to the assumption. Next consider all {x} in {|x-a| < \delta_2} and from our assumption, there exists {x} so {|f(x)-\ell |=M_3>M_2} for {|x-a|=\delta_3<\delta_2}. This keeps going and we can see that as {|x-a|} gets closer to {0}, {|f(x)-\ell |} gets larger, which contradicts to our aim that {f(x)} must approaches {\ell}. Thus, we got the second part of the definition: (more…)

LA I: Identify a list of vectors linearly dependent or linearly independent, a vector space infinite-dimensional or finite-dimensional

Inspired from some good exercises from Chapter 2A in the book Linear Algebra Done Right by Sheldon Axler.
First, let’s write out definition of linearly (in)dependent list:

Definition 1 (Linearly independence) A list of {v_1, \ldots, v_m} of vectors in {V} is called linearly independent if the only choice of {a_1, \ldots, a_m \in \mathbf{F}} so {a_1v_1+ \ldots+ a_mv_m=0} is {a_1=a_2= \ldots =a_m=0}.

The definition itself can be applied for linearly (in)dependent test. Also from this, we find

Proposition 2 Every vector in the span of linearly independet list {v_1, \ldots, v_m} can be represented uniquely as {a_1v_1+ \ldots + a_mv_m}.

Proof: Indeed, assume the contrary that {v=a_1v_1+ \ldots +a_mv_m=b_1v_1+ \ldots+b_mv_m} with {a_i} not all equal to {b_i}. Hence,

\displaystyle (a_1-b_1)v_1+(a_2-b_2)v_2+ \ldots + (a_m-b_m)v_m=0,

with {a_i-b_i} not all equal to {0}. This contradicts to our definition. \Box

This proposition doesn’t help us much in testing linearly (in)dependence. But the below proposition, which is also deduced from the definition, has better application in testing, simply by looking at the list of vectors:

Proposition 3 If some vector in a list of vectors in {V} is a linear combination of the other vectors, then the list is linearly dependent, i.e. list that is not linearly independent.

Proof: List {v_1, \ldots, v_m} has {v_1= \sum_{i}a_iv_i} then {v_1-\sum_ia_iv_i=0}, or there exists {x_1, \ldots, x_m} not all {0} so {\sum_{i=1}^mx_iv_i=0}. Hence, {v_1, \ldots, v_m} is linearly dependent. \Box


Reading a solution using algebraic integers of a HMMT number theory problem

I tried to attack the following problem but eventually gave up.

Problem 1 (HMMT Ferb 2015 Team P9) Let {z=e^{\tfrac{2\pi i}{201}}} and {\omega= e^{\tfrac{2 \pi i}{10}}}. Prove that

\displaystyle A=\displaystyle \prod_{a=0}^{9}\prod_{b=0}^{100}\prod_{c=0}^{100}(\omega^a+z^b+z^c)

is an integer and find its remainder upon division by {101}

The only idea I have is to expand this expression, but it was too complex to do. So I decided to read the solution using algebraic integers. I’m new at algebraic number theory and I don’t know much about algebraic integers. My only reference is Problems from the book (PFTB), Chapter 9.


Yellowstone Permutation – St. Petersburg MO 2015, grade 9, 2nd round

This problem was given to the St. Petersburg Olympiad 2015, 2nd round, grade 9 and is known as Yellowstone Permutation. You can see here for the article about this sequence and here for the OEIS entry.

Problem. (Yellowstone Permutation) A sequence of integers is defined as follows: a_1=1,a_2=2,a_3=3 and for n>3, a_n is the smallest integer not occurring earlier, which is relatively prime to a_{n-1} but not relatively prime to a_{n-2}. Prove that every natural number occurs exactly once in this sequence.

M. Ivanov

Here is my proof for this interesting problem. Another proof can be found in the article given link above.


The normal approximation to the binomial distribution

A coin is flipped n times with probability of getting heads is p. This is a binomial approximation X \sim B(n,p). The Year 12 MATH B textbook gives an normal approximation to this: B(n,p) \sim N(np,np(1-p))=N(\mu, \sigma^2) where \mu=E[X] is the mean (or expected value) and \sigma is the standard deviation. Since the textbook doesn’t give a proof for this so I will go and prove.


IMO 2016

IMO 2016 was held from 6/7/2016 to 16/07/2016. Although I didn’t attend this, I just want to post my solutions and how I think about the difficulty of the problems, which can be found here. I will post my solutions on this post. Many awesome solutions can be found on AoPS.


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