2
  • $\displaystyle \lim_{x\to0}\frac{\sin(\frac1x)}{\sin(\frac1x)}$
  • $\displaystyle \lim_{x\to0}\sin^{-1}\sec x$
  • $\displaystyle \lim_{x\to\pi/2}\sec^{-1}\sin x$

What I think:

  • 1
  • $\pi/2$
  • 0

But my teacher told me that limit doesn't exist for anyone of them.Why?/Why not?

Edit :See :

RE60K
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1 Answers1

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As pointed out in comments by @ThomasAndrews and @vadim123, and after I have gained more knowledge about limits I would post an answer for others who are new and might have the same problem, at first glance this may seem complex and confusing as it did to me:

  • Since $\sec x>1$ or $\sec x<1\;\forall x$. But domain of $\arcsin x$ is $[-1,1]$. So the first limit does not exist. There may be confusion that at exactly $x=0$, the domain restriction holds, but value of limit may not be equal to value of the function at that point, further, even function may be undefined there.

  • Similiarly for $\lim_{x\to0}\sin^{-1}\sec x$

  • Limit must be independent of path, if it exists. But $\displaystyle \lim_{x\to0}\frac{\sin(\frac1x)}{\sin(\frac1x)}$ has many discontinuities when one traverses the path where $\displaystyle x=\frac1{n\pi},n\in \mathbb N$[particularly of type $(0/0)$]

    Put otherwise, when one reaches the point $x=0$, you encounter infinite number of such discontinuities while traversing, moreover with increasing densities, so that the neighbourhood of $0$ is not well defined and one cannot then define the limit, or as they say, limit doesn't exist.

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RE60K
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  • You need to add bit more here. In case of $\sin(1/x)/\sin(1/x)$ the issue is not of discontinuity, but rather that the function is not defined in any neighborhood of $x = 0$. Precisely any interval around $0$ contains infinitely many points of the form $(\pm 1/n\pi)$ at which the function is not defined. – Paramanand Singh Aug 23 '14 at 06:24
  • @ParamanandSingh maybe I have written that in the last para? – RE60K Aug 23 '14 at 06:35
  • I think you have understood the thing very well for yourself but while communicating to others you have placed more focus on discontinuities rather the function being "not defined". The last para talks about "the neighborhood of $0$ not well defined". it would have been much simpler if you wrote that "the function is not defined at many points of any neighborhood of $0$". – Paramanand Singh Aug 23 '14 at 06:39
  • @ParamanandSingh I have made it community wiki, you may edit it – RE60K Aug 23 '14 at 07:12