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Math Help - Limits

  1. #1
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    Limits

    Hi. I need help in this problem. If someone could help that will be very nice. Sorry but I don't know how to write math equations on a forum post.
    Any help would be nice. Thanks

    1. Knowing that f and g are two functions and:


    lim f(x)= +oo (+infinite) and lim g(x)=0
    x->a x->a


    (a) Show that choosing conveniently the functions f and g and defining h=f*g we can have the follow situations:

    (i) lim h(x) = +oo
    x->a

    (ii) lim h(x) = -oo
    x->a

    (iii) lim h(x) = 0
    x->a

    (iv) lim h(x) = 5
    x->a

    (v) lim h(x) = -5
    x->a

    (vi) lim h(x) do not exist , without being +oo or -oo
    x->a

    Note: The point a € R is a generic point.
    €=belong
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by usual_suspect View Post
    Hi. I need help in this problem. If someone could help that will be very nice. Sorry but I don't know how to write math equations on a forum post.
    Any help would be nice. Thanks

    1. Knowing that f and g are two functions and:


    lim f(x)= +oo (+infinite) and lim g(x)=0
    x->a x->a


    (a) Show that choosing conveniently the functions f and g and defining h=f*g we can have the follow situations:
    just to clarify:

    \lim_{x \to a}f(x) = \infty and \lim_{x \to a}g(x) = 0

    and h(x) = f(x) \cdot g(x)? as opposed to h(x) = f \circ g(x)?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by usual_suspect View Post
    Hi. I need help in this problem. If someone could help that will be very nice. Sorry but I don't know how to write math equations on a forum post.
    Any help would be nice. Thanks

    1. Knowing that f and g are two functions and:


    lim f(x)= +oo (+infinite) and lim g(x)=0
    x->a x->a


    (a) Show that choosing conveniently the functions f and g and defining h=f*g we can have the follow situations:
    assuming you actually meant h(x) = f(x) \cdot g(x)

    (i) lim h(x) = +oo
    x->a
    f(x) = x^2, g(x) = \frac 1x and a = \infty



    (ii) lim h(x) = -oo
    x->a
    f(x) = -x^3, g(x) = \frac 1x and a = - \infty


    (iii) lim h(x) = 0
    x->a
    f(x) = \frac 1x, g(x) = x^2 and a = 0

    (iv) lim h(x) = 5
    x->a
    f(x) = \frac 5x, g(x) = x and a = 0

    (v) lim h(x) = -5
    x->a
    f(x) = - \frac 5x , g(x) = x and a = 0

    (vi) lim h(x) do not exist , without being +oo or -oo
    x->a
    f(x) = e^x, g(x) = \frac {\sin x}{e^x} and a = \infty


    (by the way, we can choose a right? if you're not allowed to, i leave it to you to modify my responses)
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    just to clarify:

    \lim_{x \to a}f(x) = \infty and \lim_{x \to a}g(x) = 0

    and h(x) = f(x) \cdot g(x)? as opposed to h(x) = f \circ g(x)?

    Im suposing h(x) = f(x) \cdot g(x)
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    assuming you actually meant h(x) = f(x) \cdot g(x)

    f(x) = x^2, g(x) = \frac 1x and a = \infty


    f(x) = -x^3, g(x) = \frac 1x and a = - \infty

    f(x) = \frac 1x, g(x) = x^2 and a = 0

    f(x) = \frac 5x, g(x) = x and a = 0

    f(x) = - \frac 5x , g(x) = x and a = 0

    i have to think of this one some more. i want a nice example


    (by the way, we can choose a right? if you're not allowed to, i leave it to you to modify my responses)
    Thanks Jhevon. I still don't understant how you get to the results. Can you explain a litle more.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by usual_suspect View Post
    Thanks Jhevon. I still don't understant how you get to the results. Can you explain a litle more.
    it's kind of hard to explain, you just have to "see" it

    for instance, the first. i wanted \lim_{x \to a}f(x) = \infty, \lim_{x \to a}g(x) = 0

    since i want to get \lim_{x \to a}h(x) = \infty i said to myself. how can i get the limit to go to infinty? well, i can "divide by zero" or i can have the function be, say x, or e^x or whatever such that when x goes to infinity, the function goes to infinity, so i chose the latter.

    so ok, obviously now i want \lim_{x \to \infty}, so i choose a = \infty

    now i need to get f(x) and g(x)

    well, i know that \lim_{x \to \infty} \frac 1x = 0 because i have seen that limit so many times (and you should have too). so that's my g(x), now, what do i want f(x) to be?

    i know \lim_{x \to \infty}x = \infty, so could that be f(x)? well, no, since i would have h(x) = x \cdot \frac 1x = 1 which does not go to infinity as x goes to infinity. so i realize, i just need one more factor of x to get my \lim_{x \to \infty} x = \infty, so i made f(x) = x^2. so now, \lim_{x \to \infty}f(x) = \infty, \lim_{x \to \infty}g(x) = 0 and \lim_{x \to \infty}h(x) = \lim_{x \to \infty}x^2 \cdot \frac 1x = \lim_{x \to \infty}x = \infty as desired.

    similar reasoning went into the others. just try to think of well-known limits and tweek them to fulfill the conditions. as you see, i did not use any complicated functions (well, the last one was kind of complicated, but only slightly), just keep it simple.
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    it's kind of hard to explain, you just have to "see" it

    for instance, the first. i wanted \lim_{x \to a}f(x) = \infty, \lim_{x \to a}g(x) = 0

    since i want to get \lim_{x \to a}h(x) = \infty i said to myself. how can i get the limit to go to infinty? well, i can "divide by zero" or i can have the function be, say x, or e^x or whatever such that when x goes to infinity, the function goes to infinity, so i chose the latter.

    so ok, obviously now i want \lim_{x \to \infty}, so i choose a = \infty

    now i need to get f(x) and g(x)

    well, i know that \lim_{x \to \infty} \frac 1x = 0 because i have seen that limit so many times (and you should have too). so that's my g(x), now, what do i want f(x) to be?

    i know \lim_{x \to \infty}x = \infty, so could that be f(x)? well, no, since i would have h(x) = x \cdot \frac 1x = 1 which does not go to infinity as x goes to infinity. so i realize, i just need one more factor of x to get my \lim_{x \to \infty} x = \infty, so i made f(x) = x^2. so now, \lim_{x \to \infty}f(x) = \infty, \lim_{x \to \infty}g(x) = 0 and \lim_{x \to \infty}h(x) = \lim_{x \to \infty}x^2 \cdot \frac 1x = \lim_{x \to \infty}x = \infty as desired.

    similar reasoning went into the others. just try to think of well-known limits and tweek them to fulfill the conditions. as you see, i did not use any complicated functions (well, the last one was kind of complicated, but only slightly), just keep it simple.
    ok.Thanks again. That will help very much.
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