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Math Help - 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

    f(x) = \left\{\begin{array}{ll}<br />
\frac1x, & x\text{ is rational}\\<br />
0, & x\text{ is irrational}<br />
\end{array}\right.\text.

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

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

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

    And of course, if 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:

     <br />
f(x)=\frac{\sin(x^2)}{x}<br />

    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 b>0 then for x large enough f'(x)>\epsilon>0, and so for x large enough and all y>0 (by a simple corollary to the mean value theorem):

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

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

    A similar argument applies when 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 b>0 then for x large enough f'(x)>\epsilon>0, and so for x large enough and all y>0 (by a simple corollary to the mean value theorem):

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

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

    A similar argument applies when 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
    Grand Panjandrum
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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 b>0 then for x large enough f'(x) never strays far from b, and so f'(x) is strictly greater than some positive constant (which we choose to call \epsilon) Then if we know that between x and y (with y>0) the derivetive is always greater than \epsilon>0 then: f(x+y)-f(x)>\epsilon y

    Now we are considering x as a constant, so:

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

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

    CB
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