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Q. $\lim _{x\to 0}\frac{\left(\sqrt{1-cos2x}\right)}{x}$

We can write this function as $\lim _{x\to 0}\frac{\left(\sqrt{2sin^2x}\right)}{x}$. Algebraically we have

$\frac{\left(\sqrt{2sin^2x}\right)}{x}$$=\frac{\left(\pm \sqrt{2}sinx\right)}{x}$

But as the above case is a function we are supposed to restrict the domain. When I looked into the solution of this question they consider the limit of the function as:

$\lim _{x\to 0}\frac{\left(\sqrt{2}\left|sinx\right|\right)}{x}$

When I plotted the graph of the function it gave positive values at positive 'x' and negative at negative 'x'.

And hence the limit does not exist.

But why can't we define the function as positive for all the values in the domain? Or why can't we restrict the range as positive. So that we get a symmetric graph about the y-axis rather than an inverted symmetric one? (Refer the images attached)

Am I missing any concept?

PLEASE NOTE: The graph in the second image is arbitrarily created to convey the gist of the question, the function which yields this graph has nothing to do with the question.

The actual graph looks like this

The actual graph looks like this

enter image description here Why can't it be like this?

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  • $\begingroup$ "But why can't we define the function as positive for all the values in the domain?" We can do that, sure. But that is not what the author of your text book has done. The square root symbol simply has the meaning of taking the positive root. It's not that what you want can't be done. It's just that if you do that, you are solving a different problem than the one stated. $\endgroup$ – bob.sacamento Sep 18 '14 at 18:12
  • $\begingroup$ bob.sacamento - This problem is not the creation of the author of the book I refer to, it is an official question published in one of the question papers of AIEEE(All India Engineering Entrance Examination) one of the most competitive examination in India. And the question shouldn't hold any ambiguities due to selection criteria of the examination. $\endgroup$ – Hijaz Aslam Sep 18 '14 at 18:57
  • $\begingroup$ It's someone's creation, maybe someone with authority, but still someone. They created a particular problem for you to solve. It involves $|\sin x|$ simply because the square root sign means "take the positive root". Sure, you can create another problem that would give results as in your second graph, but that's not the problem you are asked to solve. In answer to your question, "But why can't we define ... ?", well, sure, you can. But that's not what the author of the problem did. It's all a matter of meaning: The sqrt symbol means take the positive root. $\endgroup$ – bob.sacamento Sep 18 '14 at 22:51
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The answer to your question is that the numerator $\sqrt{2}|\sin{x}|$ is always positive while the denominator $x$ is negative to the left of the origin and positive to the right. Thus the quotient changes sign around the origin. I think there is some confusion about the $\pm$ in how you expressed the numerator. Before you simplify, you have $\sqrt{2\sin^2{x}}$ which is always positive. This means that the $\pm$ in your expression $\pm \sqrt{2} \sin{x}$ has to be chosen so that the expression is positive. In particular if $\sin x$ is negative then the $\pm$ has to be negative as well. There isn't a choice.

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  • $\begingroup$ Paul Sundheim - I understand that, but my question is why do we consider the function as $\sqrt{2}sinx$ in the first place rather than taking the function as positive for the 'function' sake (For instance we restrict the range of the function $\sqrt{x}$ to positive). $\endgroup$ – Hijaz Aslam Sep 18 '14 at 19:00
  • $\begingroup$ I think there is some confusion about the $\pm$ in how you expressed the numerator. Before you simplify, you have $\sqrt{2\sin^2{x}}$ which is always positive. This means that the $\pm$ in your expression $\pm \sqrt{2} \sin{x}$ has to be chosen so that the expression is positive. In particular if $\sin x$ is negative then the $\pm$ has to be negative as well. There isn't a choice. Does that help? $\endgroup$ – Paul Sundheim Sep 18 '14 at 19:18
  • $\begingroup$ Paul Sundheim - Thanks a lot. That cleared the doubt. Please do post your comment as the answer so that I can upvote it. Thanks a lot. $\endgroup$ – Hijaz Aslam Sep 18 '14 at 19:51

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