Answer
The limit does not exist.
Work Step by Step
\begin{align*} \lim _{t \rightarrow 0^+}\frac{\sqrt{1-\cos t}}{t}&= \lim _{t \rightarrow 0^+}\left(\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=\left( \lim _{t \rightarrow 0^+}\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=\left(\frac{1}{2} \right)^{1/2}\\ &=\frac{1}{\sqrt{2}}. \end{align*}
Where we used the fact that $\lim _{x \rightarrow 0^+} \frac{1-\cos x}{x^2}=\frac{1}{2}$. By the same way, we have
\begin{align*} \lim _{t \rightarrow 0^-}\frac{\sqrt{1-\cos t}}{t}&=- \lim _{t \rightarrow 0^-}\left(\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=-\left( \lim _{t \rightarrow 0^-}\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=-\left(\frac{1}{2} \right)^{1/2}\\ &=-\frac{1}{\sqrt{2}}. \end{align*}
Hence $ \lim _{t \rightarrow 0^+}\frac{\sqrt{1-\cos t}}{t}\neq \lim _{t \rightarrow 0^-}\frac{\sqrt{1-\cos t}}{t}$ and so the limit does not exist.
