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Math Help - L'Hospital's Rule True/False

  1. #1
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    L'Hospital's Rule True/False

    Hi, can anyone tell me which of these are wrong?

    1. The indeterminate form of type "0/0" can approach any real number as a limit -- False

    2. There is a natural number n such that the power function f(x)=x^n is growing faster than the exponential function g(x)=e^x -- True

    3. if lim x->infinity f(x) = +infinity, then for any k (all real numbers), lim x->+infinity k/f(x) = 0 -- False

    4. The function f(x) = lnx grows more slowly than any positive power of x -- True

    Thanks!
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  2. #2
    Moo
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    Quote Originally Posted by coldfire View Post
    Hi, can anyone tell me which of these are wrong?

    1. The indeterminate form of type "0/0" can approach any real number as a limit -- False
    ok

    3. if lim x->infinity f(x) = +infinity, then for any k (all real numbers), lim x->+infinity k/f(x) = 0 -- False
    not ok
    why do you say it's false ?

    2. There is a natural number n such that the power function f(x)=x^n is growing faster than the exponential function g(x)=e^x -- True

    4. The function f(x) = lnx grows more slowly than any positive power of x -- True
    the exponential grows faster than any power of x, which grows faster than the logarithm.
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  3. #3
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    disagree with #1 if the limit exists.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by coldfire View Post
    Hi, can anyone tell me which of these are wrong?
    1. The indeterminate form of type "0/0" can approach any real number as a limit -- False
    True, think about \lim_{x\to{0}}\frac{1-\cos(x)}{x^2}

    This yields \frac{1}{2}

    So if a\in\mathbb{R} then \lim_{x\to{0}}\frac{a-a\cos(x)}{x^2}=\frac{a}{2}

    Supposing the limit exists of course.
    2. There is a natural number n such that the power function f(x)=x^n is growing faster than the exponential function g(x)=e^x -- True
    No! Consider that the function x^n can be differentiated and n+1 number of times to give n! and if you differentiate e^x an n+1 number of times you get e^x and it is pretty obvious that

    \forall{n}\in\mathbb{N}~\exists{N}\backepsilon\for  all{x}\geqslant{N}~n!\leqslant{e^x}
    3. if lim x->infinity f(x) = +infinity, then for any k (all real numbers), lim x->+infinity k/f(x) = 0 -- False
    True! \begin{aligned}\lim_{x\to\infty}\frac{k}{f(x)}&=k\  lim_{x\to\infty}\frac{1}{f(x)}\\<br />
&=k\cdot{0}\left({\color{red}\text{Why?}}\right  )\\<br />
&=0\end{aligned}
    4. The function f(x) = lnx grows more slowly than any positive power of x -- True
    Yes! The reason being that \left(\ln(x)\right)'=\frac{1}{x}\leqslant{\varphi{  x}^{\varphi-1}}~\varphi>0
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  5. #5
    Moo
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    Quote Originally Posted by skeeter View Post
    disagree with #1 if the limit exists.
    Quote Originally Posted by Mathstud28 View Post
    True
    What's disturbing me for the first one is that it is written "can reach any real value"

    so what I understand is that since it is an indeterminate form 0/0, we can attribute any value we want to the limit. Which is not true.
    Can someone explain it more clearly ?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    What's disturbing me for the first one is that it is written "can reach any real value"

    so what I understand is that since it is an indeterminate form 0/0, we can attribute any value we want to the limit. Which is not true.
    Can someone explain it more clearly ?
    Does this help? Suppose \lim_{x\to{c}}\frac{f(x)}{f(c)}=L where f(c)=g(c)=0

    Then the limit \lim_{x\to{c}}\frac{\gamma{f(x)}}{g(x)}=\gamma{L}~  \gamma\in\mathbb{R}

    Note this would still be a "0/0" form since \gamma{f(c)}=\gamma{0}=0~\gamma\in\mathbb{R}

    Now let \varphi\in\mathbb{R}, let this be the "any value" desired, then all thats left to do is solve \gamma{L}=\varphi
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