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Thread: limits of functions

  1. #1
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    limits of functions

    Im struggling with a couple of questions... any help would be appreciated. You have to answer true or false and if true give a proof, if false give a counterexample.

    (a) If lim(f(x)) as x tends to infinity is finite and limf'(x)=b then b=0.

    I think this is true but i have no idea how to prove it, I wouldn't even know where to start!

    (b) If lim(f(x)) as x tends to infinity is finite, the lim(f'(x))=0.

    I think this must be false because of the subtle difference in the question - is it false because the limit of the derivative function might not actually exist? This seems to make sense to me as the only way it could be false, although I cannot think of a counterexample.

    thanks
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    Quote Originally Posted by claire511 View Post
    (a) If lim(f(x)) as x tends to infinity is finite and limf'(x)=b then b=0.

    (b) If lim(f(x)) as x tends to infinity is finite, the lim(f'(x))=0.
    For the second, here is a counterexample: Consider

    $\displaystyle f(x) = \left\{\begin{array}{ll}
    \frac1x, & x\text{ is rational}\\
    0, & x\text{ is irrational}
    \end{array}\right.\text.$

    Now, $\displaystyle \lim_{x\to\infty}f(x)=0,$ and is thus finite, which is easy to show by the squeeze/sandwich theorem, using $\displaystyle x\mapsto\frac1x$ and $\displaystyle x\mapsto0\text.$

    But, $\displaystyle f(x)$ is nowhere continuous. To show this, choose any $\displaystyle c\in\mathbb R\text.$

    First, if $\displaystyle c$ is rational, then $\displaystyle f(c) = \frac1c$. Take $\displaystyle \varepsilon=\frac1{2c}\text.$ Now, no matter how small you choose $\displaystyle \delta>0$ to be, there is always going to be an irrational number $\displaystyle d$ within $\displaystyle \delta$ units of $\displaystyle c,$ in which case $\displaystyle \left\lvert f(d)-\frac1c\right\rvert=\left\lvert0-\frac1c\right\rvert=\frac1c>\frac1{2c}=\varepsilon \text.$ Thus, $\displaystyle \lim_{x\to c}f(x)\neq f(c)$ and $\displaystyle f$ is not continuous at $\displaystyle c\text.$ The case for irrational $\displaystyle c$ should be similar.

    And of course, if $\displaystyle f$ is everywhere discontinuous, then at any given value its derivative is undefined.

    Does that help any? I could probably get you started on (a), but you'll have to wait a bit.
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  3. #3
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    Quote Originally Posted by claire511 View Post

    (b) If lim(f(x)) as x tends to infinity is finite, the lim(f'(x))=0.
    Consider:

    $\displaystyle
    f(x)=\frac{\sin(x^2)}{x}
    $

    CB
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  4. #4
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    Quote Originally Posted by claire511 View Post

    (a) If lim(f(x)) as x tends to infinity is finite and limf'(x)=b then b=0.
    If $\displaystyle b>0$ then for $\displaystyle x$ large enough $\displaystyle f'(x)>\epsilon>0$, and so for $\displaystyle x$ large enough and all $\displaystyle y>0$ (by a simple corollary to the mean value theorem):

    $\displaystyle f(x+y)>f(x)+\epsilon y$

    hence the limit of this as $\displaystyle y \to \infty$ cannot be finite, which contradicts the assumption that $\displaystyle \lim_{x \to \infty}f(x)$ is finite.

    A similar argument applies when $\displaystyle b<0$.

    Now you will need to fill in the detail to complete this.

    CB
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    Quote Originally Posted by CaptainBlack View Post
    If $\displaystyle b>0$ then for $\displaystyle x$ large enough $\displaystyle f'(x)>\epsilon>0$, and so for $\displaystyle x$ large enough and all $\displaystyle y>0$ (by a simple corollary to the mean value theorem):

    $\displaystyle f(x+y)>f(x)+\epsilon y$

    hence the limit of this as $\displaystyle y \to \infty$ cannot be finite, which contradicts the assumption that $\displaystyle \lim_{x \to \infty}f(x)$ is finite.

    A similar argument applies when $\displaystyle b<0$.

    Now you will need to fill in the detail to complete this.

    CB
    Thank you so much for replying. I understand by corollary to MVT if y positive f(x+y)>f(x) but why is there a plus epsilon times y bit on the end :S?? then how does letting y tend to infinity show that the limit of x tending to infinity of f(x) is not finite?
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  6. #6
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    Quote Originally Posted by claire511 View Post
    Thank you so much for replying. I understand by corollary to MVT if y positive f(x+y)>f(x) but why is there a plus epsilon times y bit on the end :S?? then how does letting y tend to infinity show that the limit of x tending to infinity of f(x) is not finite?
    If $\displaystyle b>0$ then for $\displaystyle x$ large enough $\displaystyle f'(x)$ never strays far from $\displaystyle b$, and so $\displaystyle f'(x)$ is strictly greater than some positive constant (which we choose to call $\displaystyle \epsilon$) Then if we know that between $\displaystyle x$ and $\displaystyle y$ (with $\displaystyle y>0$) the derivetive is always greater than $\displaystyle \epsilon>0$ then: $\displaystyle f(x+y)-f(x)>\epsilon y$

    Now we are considering $\displaystyle x$ as a constant, so:

    $\displaystyle \lim_{u \to \infty} f(u)=\lim_{y \to \infty} f(x+y) $

    but we see from earlier that $\displaystyle \lim_{y \to \infty} f(x+y) =\infty$

    CB
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