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Thread: need help with three conceptual calculus problems!

  1. #1
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    need help with three conceptual calculus problems!

    ok so 1.

    true or false (and provide reasoning, not a full length proof) that
    lim as x -> a of f(x)*g(x) can not exist
    if lim x->a f(x) and lim x->a g(x) both exist
    so its kind of a reasoning to show that the product limit law always works if both limits exist?


    2. In the formal definition of limits: lim x->c f(x) = L, if for every E>0 there is a d>0 such that IF 0<|x-c|<d THEN |f(x)-L|<E

    can you switch the inequalities after the IF and THEN, and does the definition still hold?


    3. TRUE or FALSE

    f(x) = 1 if x is rational
    = 2 if x is irrational

    the limit as X-->0 exists for f(x)

    im not sure about this one cause of whether the right and left hand limits meet and if so, at 1 or 2??!?!


    thanks! please do asap
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  2. #2
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    Quote Originally Posted by stones44 View Post
    true or false (and provide reasoning, not a full length proof) that
    lim as x -> a of f(x)*g(x) can not exist
    if lim x->a f(x) and lim x->a g(x) both exist
    so its kind of a reasoning to show that the product limit law always works if both limits exist?
    If $\displaystyle \lim_{x\to a}f(x)$ and $\displaystyle \lim_{x\to a}g(x)$ exist then $\displaystyle \lim_{x\to a}f(x)g(x)$ exists. That is a theorem.

    2. In the formal definition of limits: lim x->c f(x) = L, if for every E>0 there is a d>0 such that IF 0<|x-c|<d THEN |f(x)-L|<E

    can you switch the inequalities after the IF and THEN, and does the definition still hold?
    No

    3. TRUE or FALSE

    f(x) = 1 if x is rational
    = 2 if x is irrational

    the limit as X-->0 exists for f(x)

    im not sure about this one cause of whether the right and left hand limits meet and if so, at 1 or 2??!?!
    Assume that $\displaystyle \lim_{x\to 0}f(x) = L$ (that is, assume it exists).


    If $\displaystyle L < 1$ then pick $\displaystyle \epsilon > 0$ so that $\displaystyle L + \epsilon < 1$.
    This means there is $\displaystyle \delta > 0$ so that if $\displaystyle 0 < |x| < \delta \implies |f(x) - L| < \epsilon$.
    Therefore $\displaystyle L - \epsilon < f(x) < L + \epsilon < 1$ for $\displaystyle 0 < |x| < \delta$.
    But this is a contradiction since $\displaystyle f(x) \not < 1$.

    If $\displaystyle L > 2$ then pick $\displaystyle \epsilon > 0$ so that $\displaystyle L - \epsilon > 2$.
    Repeating a similar argument will show $\displaystyle f(x) > 2$ and this is a contradiction too.

    Therefore $\displaystyle 1\leq L \leq 2$ if it exists. Using a similar argument as above we can show $\displaystyle L\not= 1, L\not = 2$. This forces $\displaystyle 1 < L < 2$. But then this means there is $\displaystyle \epsilon > 0$ so that $\displaystyle 1< L - \epsilon$ and $\displaystyle L+\epsilon < 2$.
    Thus, there is $\displaystyle \delta > 0$ so that $\displaystyle 0 < |x| < \delta \implies |f(x) - L| < \epsilon$.
    Thus, $\displaystyle 1 < L - \epsilon < f(x) < L + \epsilon < 2$.
    This is a contradiction beceause $\displaystyle f$ does not take values strictly between $\displaystyle 1$ and $\displaystyle 2$.
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