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Math Help - Finding functions by a given Limits!

  1. #1
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    Question Finding functions by a given Limits!

    Hello I gotta find functions by a given limits ;

    if there is no such functions just tell me by sayin no such function

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    f(x):={x+b(x!=0),17(x=0)}

    What values of b give values for the limit of f(x) (x==>0) = L

    Need b , and L ( Limit)

    b doesnt give L for any value of it
    b gives L for any value of it
    b give L for only b>0
    b give L for only b<0
    b give L for only b=0
    b give L for only b=17
    b give L for only b=-17
    b=?,L=? (give L= ! when there is no b )

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    Need a function that
    limit (x=>zero) F(x) = infinity
    limit x(=>zero) X*F(x) = zero
    -----

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    Need a function that
    limit (x=>infinity) F(x) = 0
    limit x(=> 0 - ) F(x) = 3
    limit x(=> 0 + ) F(x) =1
    This function can be written in two ways :
    for x<0
    or x>= 0
    -----
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  2. #2
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    I recommend you go back and re-read the problem. What you wrote here, makes no sense. \lim_{x\to 0} x+ b= b so you make that limit equal to "L" by taking b= L. But then b= L can be any number. Did the problem say something about "continuous"?

    \lim_{x\to 0} F(x)= \infty, \lim_{x\to 0} xF(x)= 0. So F must go to infinity as x goes to 0 but not as fast as x itself. How about 1 over x to any power less than 1? Say, 1/\sqrt{x}= x^{-1/2}?

    \lim_{x\to \infty} F(x)= 0, \lim_{x\to 0^-}F(x)= 3, \lim_{x\to 0^+}F(x)= 1

    As you say, do this as two separate problems. First find a function that goes to 0 as x goes to infinity and goes to 1 as x goes to 0. There are many such functions. A fraction with denominator of higher degree than numerator but with constant terms in numerator and denominator the same with work. Then find a function that goes to 3 as x goes to 0 (for x< 0 but if you find one that works for x> 0, just replace x with -x).
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I recommend you go back and re-read the problem. What you wrote here, makes no sense. \lim_{x\to 0} x+ b= b so you make that limit equal to "L" by taking b= L. But then b= L can be any number. Did the problem say something about "continuous"?

    \lim_{x\to 0} F(x)= \infty, \lim_{x\to 0} xF(x)= 0. So F must go to infinity as x goes to 0 but not as fast as x itself. How about 1 over x to any power less than 1? Say, 1/\sqrt{x}= x^{-1/2}?

    \lim_{x\to \infty} F(x)= 0, \lim_{x\to 0^-}F(x)= 3, \lim_{x\to 0^+}F(x)= 1

    As you say, do this as two separate problems. First find a function that goes to 0 as x goes to infinity and goes to 1 as x goes to 0. There are many such functions. A fraction with denominator of higher degree than numerator but with constant terms in numerator and denominator the same with work. Then find a function that goes to 3 as x goes to 0 (for x< 0 but if you find one that works for x> 0, just replace x with -x).
    Hey there,first thank for the help.I couldnt understand what u meant in 1 + 3 but as for 2 its true (1/x^1/2) for second condition ( when multiplaied with x) but not for the first!
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