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
Why can't it be like this?
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.