Q.$\displaystyle\lim_{x\to 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)$

Using $1-\cos x=2\sin^2(x/2)$, $\sqrt{1-\cos x}=\sqrt2\left|\sin(x/2)\right|$. For $x\to0^+$, $\sin(x/2)>0$ so $\left|\sin(x/2)\right|=\sin(x/2)$, giving $\dfrac{\sqrt2\sin(x/2)}{x}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sin(x/2)}{x/2}\to\dfrac{\sqrt2}{2}$. For $x\to0^-$, $\sin(x/2)<0$ so $\left|\sin(x/2)\right|=-\sin(x/2)$, giving $\dfrac{-\sqrt2\sin(x/2)}{x}\to-\dfrac{\sqrt2}{2}$. Since the right-hand limit $\left(\tfrac{\sqrt2}2\right)$ and left-hand limit $\left(-\tfrac{\sqrt2}2\right)$ disagree, the two-sided limit does not exist — the presence of the square root (which is always …

Rewrite $1-\cos x$ via the half-angle identity, note that the square root forces an absolute value, and evaluate the two one-sided limi …

Dropping the absolute value around $\sin(x/2)$ and reporting a single value $1/\sqrt2$ as if the limit existed two-sidedly — an easy trap since $\sqrt{\cdot}$ is o …