Limits of specific functions 
*

*$\displaystyle \lim_{x\to0}\frac{\sin(\frac1x)}{\sin(\frac1x)}$

*$\displaystyle \lim_{x\to0}\sin^{-1}\sec x$

*$\displaystyle \lim_{x\to\pi/2}\sec^{-1}\sin x$


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 :


*

*$\displaystyle \lim_{x\to0}\frac{\sin(\frac1x)}{\sin(\frac1x)}$

*$\displaystyle \lim_{x\to0}\sin^{-1}\sec x$

*$\displaystyle \lim_{x\to\pi/2}\sec^{-1}\sin x$

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.
