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What a weird function.

I tried to find out: $$\lim_{x\to\infty } \frac{\ln x}{\sqrt{x}\,{\sin{x}}}$$

So, I can't use L'Hopital 'cause there's no actual limit in the denominator. It doesn't exist.

Then, I tried to use Heine's theorem and chose two sequences, but yet I got the same limit.

I believe it does not converge. How can I prove it?

Thanks

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3 Answers 3

up vote 11 down vote accepted

You can consider a sequence $x_n = \pi n - 2^{-n}$. On this sequence your function will become unbounded while this sequence goes to infinity. Hence the limit does not exist.

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I don't get your point, can you please extend your answer? –  user6163 Jun 22 '11 at 18:05
4  
A divergent subsequence implies... –  The Chaz 2.0 Jun 22 '11 at 18:10
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In more informal terms: the funcion goes to (plus/minus) infinity at $\pi, 2 \pi, 3\pi \cdots$ . Hence, it cannot have a limit. –  leonbloy Jun 22 '11 at 18:14

The limit doesn't exists, since the denominator has infinitely many zeros so you cannot find an N such that for $x\ge N$, $f$ is well-defined.

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For the limit to exist, it's not necessary for $f$ to be defined for all $x$ beyond some point. For example, if $f=1$ everywhere except that $f$ is undefined for all integers, then $\lim_{x\to\infty} f(x)=1$ (according to the definition of limit in Rudin's Principles (Def. 4.33), for example). –  Hans Lundmark Jun 22 '11 at 21:14
    
@Hans Lundmark if you consult with Wikipedia en.wikipedia.org/wiki/Limit_of_a_function then it is required that $f$ to be defined for all $x$ beyond some point. Anyway, different definitions. –  timhortons Jun 23 '11 at 14:15

Still not enough rep to comment. Hardy's "Pure Mathematics" does as good a job as any book I've ever read at analysing the different kinds of behaviour of such sequences, with many examples.

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