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Math Help - confuse.....

  1. #1
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    Unhappy confuse.....

    what is limit? how to link limit with differentiation? how to link limit with the first principle when i learnt in form 5? how they work out the first principle?
    question :
    1) modulus X+4 smaller or equal to modulus X-3.
    why can we square both side? why is square so powerful? we can use it to solve so much things.why?
    2) the differentiation of exponetial come up with a few formula.can omebody prove me how to get all this equation?
    y=e( to the power of mx)
    dy/dx=e(to the power of mx) multiply with d (mx)/dx
    why?????
    i'm an ausmat student. pls somebody help me.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by yong View Post
    what is limit? how to link limit with differentiation? how to link limit with the first principle when i learnt in form 5? how they work out the first principle?
    question :
    1) modulus X+4 smaller or equal to modulus X-3.
    why can we square both side? why is square so powerful? we can use it to solve so much things.why?
    2) the differentiation of exponetial come up with a few formula.can omebody prove me how to get all this equation?
    y=e( to the power of mx)
    dy/dx=e(to the power of mx) multiply with d (mx)/dx
    why?????
    i'm an ausmat student. pls somebody help me.
    before that, what do you know so far? i mean, what have the school taught you? have they introduced you to the concepts of limits?
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  3. #3
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    thanks for your fast reply.it is staggering.i know how to do the question. but i dun know what is the point to learn this.and i need to know the linkn between differentiation and limit. i just know how to fill in limit into the question.eg. when limit x approaches zero or approaches infinity.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by yong View Post
    ... link between differentiation and limit. ..
    ok.. do you know the limit definition of differentiation? what was the definition of differentiation they gave you?
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  5. #5
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    erm.i dun know the meaning of limit to differentiation. and they didn't give us the meaning of differentiation also. my hostel dun have wifi.so i maybe late to reply.sorry.thanks for your help also
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  6. #6
    MHF Contributor kalagota's Avatar
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    so you are trying to study topics in mathematics on your own.. that is a good habit, (at least for me)!

    well, i don't have lots of time today so I'll just state some.. however, i will assume that you know at least what a function is, the domain and range, kinds of functions, etc

    so, we will start with these assumptions:
    X \subseteq \mathbb{R}
    f: X \rightarrow \mathbb{R} (this means that our function is in terms of one variable only, and in particular real-valued function)
    a \in X

    INTUITIVELY:
    The limit of the function f(x) as x approaches a is L, written \lim_{x\rightarrow a} f(x) = L, means that the values of f(x) gets closer and closer to L as the values of x gets closer and closer to a (but not necessarily equal to a).


    Here, we say that we can make f(x) as close as we like to L by taking values of x that are sufficiently close to a (but not necessarily equal to a).

    EXAMPLE: Let f(x) = 2x + 1. We want to find the limit of f(x) as x approaches to 1.

    So, using that intuitive "definition", let us take values of x which are close to 1 but not equal to 1.

    \begin{array}{c|cccccccc} x & 0 & 0.5 & 0.8 & 0.9 & 0.99 & 0.999 &  0.9999 & ....... \\  f(x) &  1 & 2 & 2.6 & 2.8 & 2.98 & 2.998 & 2.9998 & ....... \end{array}

    \begin{array}{c|cccccccc} x &     2  &   1.5 &    1.3   &  1.1  &   1.01   &  1.0001 &   1.00000001 & ....... \\ f(x) &   5 &      4  &    3.6   &  3.2&     3.02 &    3.0002   & 3.00000002 & ....... \end{array}

    Now, observe that as we take values of x close to 1, we see that the values of f(x) gets close to 3.
    So "intuitively", we say that the limit of f(x) as x approaches to 1 is 3, that is, \lim_{x\rightarrow 1} f(x) = \lim_{x\rightarrow 1} 2x + 1 = 3..

    as an easy exercise, can you tell the answer to the next question:
    \lim_{x\rightarrow 2} \frac{x^2 - 4}{x-2}


    if no other helpers will post, my next post will be the "true" definition of limit, i.e. the definition using delta-epsilon.
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  7. #7
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    ok.this helps me a lot.till now i still can catch you.i'm still with you.but then the last thing you mentioned....something about delta-epsilon. that thing i haven't heard before. so when you exlpain next time can you give it in detail? it maybe troublesome.sorry for that.
    thanks for your help anyway,really.i want to know more.more.and more.i'll be waiting.
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  8. #8
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    sorry.

    oh my god.i forgot. the answer is probably 4 is it?thanks...
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  9. #9
    MHF Contributor kalagota's Avatar
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    Q: when you say a is a non negative number and a<x for any positive real number, what can you conclude about a, i mean, what is a(the) possible value for a? (give me an answer..)

    Definition: Let a\in X\subseteq\mathbb{R} and let f: X \rightarrow \mathbb{R} be a function. We say that limit of f(x) as x approaches a is L, denoted by \lim_{x\rightarrow a} f(x) = L if and only if for every (epsilon) \varepsilon > 0, there is a (delta) \delta>0 such that |f(x) - L| < \varepsilon whenever 0<|x-a|<\delta.

    Remark: We shall assume that L<\infty, that is, L exists.
    Remark: The \delta is usually in terms of \varepsilon, that is \delta is a function of \varepsilon. Hence, \delta depends on the value of \varepsilon.

    So how do we use this?
    Let us take the first example:
    f(x)=2x+1

    By our first method, we concluded that the limit of f(x) is 3 as x goes to 1, and we had that conclusion by observing the behavior of f(x). The formal way to prove it is using the epsilon-delta definition.

    Proof:
    So, we let \varepsilon >0 be an arbitrary positive real number. We want to find a necessary \delta>0 (in terms of epsilon) such that if 0<|x-1|<\delta, then |f(x) - 3| < \varepsilon.

    Let us take a look on |f(x) - L|.
    |f(x) - L| = |2x+1 - 3| = |2x - 2| = |2||x - 1| =2|x - 1| and we want it to be less than epsilon, that is 2|x - 1| < \varepsilon.

    So let us take \delta = \frac{\varepsilon}{2}. Then whenever 0<|x-1|<\delta=\frac{\varepsilon}{2}, we have |f(x) - L| = 2|x - 1| < 2\delta = 2 \cdot \frac{\varepsilon}{2} = \varepsilon . Therefore \lim_{x\rightarrow 1} 2x+1 = 3.


    Can you still follow? Just raise your questions if you have any.
    And also, prove that \lim_{x\rightarrow2}\frac{x^2-4}{x-2} = 4.
    Last edited by kalagota; August 8th 2008 at 07:08 PM.
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  10. #10
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    A limit is the value of a function as its variable approaches a certain number.
    For example, take y=x, or f(x) = 1/x. If we create a table we see the following:

    x f(x)
    1 1
    10 1/10
    100 1/100

    This will continue until infinity. A limit tries to describe what the behavior of the graph is once it "reaches" infinity. In this case, you can see the value of f(x) become smaller as x increases. In math, we say that the limit of f(x) as x approaches infinity is 0. You can think of it as f(infinity) = 0, but this is not entirely correct.

    Differentiation uses limits to find the slope of a given point. Without calculus, the only way to find a slope is to take 2 points, and use the slope formula (f(x)2-f(x)1)/(x2-x1). Now, imagine that the distance between these two points becoming infinately smaller.

    Im afraid the rest of your questions dont really make sense to me, but hopefully this helps.
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  11. #11
    MHF Contributor kalagota's Avatar
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    Now, before we go to what Dubulus discussed, let us take a look on some properties of limits.

    A. Suppose \lim_{x\rightarrow a} f(x) exists. Then this limit is unique, i.e. if \lim_{x\rightarrow a} f(x) = L_1 and \lim_{x\rightarrow a} f(x) = L_2, then it must be L_1 = L_2.

    B. The limit of a constant function is the constant itself, i.e. \lim_{x\rightarrow a} c = c.

    C. Suppose \lim_{x\rightarrow a} f(x) = L_1 and \lim_{x\rightarrow a} g(x) = L_2. Then,

    1. \lim_{x\rightarrow a} \left(f(x)+g(x)\right) = L_1 + L_2
    2. \lim_{x\rightarrow a} \left(f(x)-g(x)\right) = L_1 - L_2
    3. \lim_{x\rightarrow a} \left(f(x)g(x)\right) = L_1 \cdot L_2
    4. \lim_{x\rightarrow a} \left(\frac{f(x)}{g(x)}\right) = \frac{L_1}{L_2} if g(a) \not=0 and L_2\not=0.

    other properties can be deduced from here.

    Remark: If f(x) is a function for which f(a) exists, then \lim_{x\rightarrow a} f(x) = f(a).

    Theorems.
    1. \lim_{x\rightarrow a} f(x) = L \Longleftrightarrow \lim_{x\rightarrow a} f(x) - L = 0
    2. \lim_{x\rightarrow a} f(x) = L \Longleftrightarrow \lim_{t\rightarrow 0} f(t + a) = L


    well, before i proceed, i have to know whether you still can catch up.
    Last edited by kalagota; August 10th 2008 at 03:26 AM.
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  12. #12
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    Quote Originally Posted by kalagota View Post
    Q: when you say a is a non negative number and a<x for any positive real number, what can you conclude about a, i mean, what is a(the) possible value for a? (give me an answer..)

    Definition: Let a\in X\subseteq\mathbb{R} and let f: X \rightarrow \mathbb{R} be a function. We say that limit of f(x) as x approaches a is L, denoted by \lim_{x\rightarrow a} f(x) = L if and only if for every (epsilon) \varepsilon > 0, there is a (delta) \delta>0 such that |f(x) - L| < \varepsilon whenever 0<|x-a|<\delta.

    Remark: We shall assume that L<\infty, that is, L exists.
    Remark: The \delta is usually in terms of \varepsilon, that is \delta is a function of \varepsilon. Hence, \delta depends on the value of \varepsilon.

    So how do we use this?
    Let us take the first example:
    f(x)=2x+1

    By our first method, we concluded that the limit of f(x) is 3 as x goes to 1, and we had that conclusion by observing the behavior of f(x). The formal way to prove it is using the epsilon-delta definition.

    Proof:
    So, we let \varepsilon >0 be an arbitrary positive real number. We want to find a necessary \delta>0 (in terms of epsilon) such that if 0<|x-1|<\delta, then |f(x) - 3| < \varepsilon.

    Let us take a look on |f(x) - L|.
    |f(x) - L| = |2x+1 - 3| = |2x - 2| = |2||x - 1| =2|x - 1| and we want it to be less than epsilon, that is 2|x - 1| < \varepsilon.

    So let us take \delta = \frac{\varepsilon}{2}. Then whenever 0<|x-1|<\delta=\frac{\varepsilon}{2}, we have |f(x) - L| = 2|x - 1| < 2\delta = 2 \cdot \frac{\varepsilon}{2} = \varepsilon . Therefore \lim_{x\rightarrow 1} 2x+1 = 3.


    Can you still follow? Just raise your questions if you have any.
    And also, prove that \lim_{x\rightarrow2}\frac{x^2-4}{x-2} = 4.

    i dun know the answer. can you tell me how to do pls? and actually i dun understand the delta -epsilon theory even a bit. sorry. can you explain it in simpler way? i'm a bit stupid. sorry.
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  13. #13
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    Smile

    Quote Originally Posted by kalagota View Post
    Now, before we go to what Dubulus discussed, let us take a look on some properties of limits.

    A. Suppose \lim_{x\rightarrow a} f(x) exists. Then this limit is unique, i.e. if \lim_{x\rightarrow a} f(x) = L_1 and \lim_{x\rightarrow a} f(x) = L_2, then it must be L_1 = L_2.

    B. The limit of a constant function is the constant itself, i.e. \lim_{x\rightarrow a} c = c.

    C. Suppose \lim_{x\rightarrow a} f(x) = L_1 and \lim_{x\rightarrow a} g(x) = L_2. Then,

    1. \lim_{x\rightarrow a} \left(f(x)+g(x)\right) = L_1 + L_2
    2. \lim_{x\rightarrow a} \left(f(x)-g(x)\right) = L_1 - L_2
    3. \lim_{x\rightarrow a} \left(f(x)g(x)\right) = L_1 \cdot L_2
    4. \lim_{x\rightarrow a} \left(\frac{f(x)}{g(x)}\right) = \frac{L_1}{L_2} if g(a) \not=0 and L_2\not=0.

    other properties can be deduced from here.

    Remark: If f(x) is a function for which f(a) exists, then \lim_{x\rightarrow a} f(x) = f(a).

    Theorems.
    1. \lim_{x\rightarrow a} f(x) = L \Longleftrightarrow \lim_{x\rightarrow a} f(x) - L = 0
    2. \lim_{x\rightarrow a} f(x) = L \Longleftrightarrow \lim_{t\rightarrow 0} f(t + a) = L


    well, before i proceed, i have to know whether you still can catch up.

    this one i can accept. i think i know how to do. but except the final one. the second theorem. can you explain that to me?thanks.
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  14. #14
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by yong View Post
    i dun know the answer.
    the only possible value for a is 0.

    we have the assumption that x is any positive number, a is non-negative such that a<x.

    now, let us take a=1. this is a nonnegative number. but since x is any positive number, x can be .5 but this does not satisfy a<x.
    now, let us take a=.1 and this is a nonnegative number. again, x can be any positive number so i can take x=.09. and again, a<x is not satisfied.
    so if you do this continuously, you will arrive at a=0 a non negative number that satisfy a<x for any positive number x.

    Quote Originally Posted by yong View Post
    can you tell me how to do pls? and actually i dun understand the delta -epsilon theory even a bit. sorry. can you explain it in simpler way? i'm a bit stupid. sorry.
    no, not really. your not stupid. it is normal to for beginners.. anyways, it is hard to explain it in this way, so i'll just let you watch this,
    although...
    i hope you can understand what the speaker said.. but keep in mind that the f(a) he was talking may not exist, i.e. if you concentrate very well, he will tell you we are just trying to approach a we really don't get to a exactly, (as what the intuitive definition always tells us..)

    if you still have questions on that, fell free to ask.
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  15. #15
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by yong View Post
    this one i can accept. i think i know how to do. but except the final one. the second theorem. can you explain that to me?thanks.
    see this.. x\rightarrow a \Longrightarrow x-a\rightarrow a-a \Longrightarrow x-a\rightarrow 0

    so we can set t=x-a. this means that t+a=x..

    from here, can guess what will happen next?
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