One thought says that since the limit x tends to 1+ does not exist therefore the limit does not exist. But other thought says that since the function is not defined at 1+, i.e. it is not in the domain of the function so we must not discuss limit over there and the limit is by default the limit x tends to 1-. Which thought is correct?
Does limit x tends to 1 sin inverse x exist?
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$\begingroup$
limits
trigonometry
inverse-function
3 Answers
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$\text{Arcsin} \ x$ is defined on the interval $[-1,1]$, so you cannot take a right hand limit to $1$. So to answer your question directly, the latter thought you proposed is more accurate.
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0I corrected the post to try to clarify my doubt. My second thought meant that the limit exists as limit x→1 for arcsinx is limit x→1- only. – 2017-01-15
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0If you restrict yourself to the domain of the function, then $\lim_{x \to 1}$ is $\lim_{x \to 1^-}$, simply because there is no way to approach $1$ from the right hand, within your domain. – 2017-01-15
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Are you talking about "$\lim_\limits{x\to1^{+}}\arcsin x$"? Since the function is undefined for $x>1$, this is a moot question. On the other hand, your both thoughts (some of which isn't quite clear) end up with the same conclusion that this limit doesn't exist. So I'm not sure what your question is, if there's no disagreement to resolve.
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$\lim_{x\to 1}\sin^{-1}(x)$ does not exist, because $\sin^{-1}(x)$ is not defined for $x>1$. But $\lim_{x\to 1^-}\sin^{-1}(x)$ does exist, and is equal to $\sin^{-1}(1)=\pi/2$.