Does L'Hopital's work here?

using expansion cos(x) = 1-(x^2)/2!+(x^4)/4!-.....so on 1-cos(x)=(x^2)/2!-(x^4)/4!+....so on then just take x^2 common from all terms and apply limit

$\lim_{x \rightarrow 0^{+}} \dfrac{x}{\sqrt{1-\cos x}} \cdot \dfrac{\sqrt{1+\cos x}}{\sqrt{1+ \cos x}}$ $= \lim_{x \rightarrow 0^{+}} \dfrac{x \sqrt{1+\cos x}}{\sqrt{1-\cos^{2} x}}$ $= \lim_{x \rightarrow 0^{+}} \dfrac{x \sqrt{1+\cos x}}{\left| \sin x \right|}$ We know that $\displaystyle \lim_{x \rightarrow 0} \dfrac{\sin x}{x} = 1$ and it's fine if it's flipped because the numerator in that key limit is approximately the same as the denominator when x approaches zero. $= \lim_{x \rightarrow 0^{+}} \sqrt{1+\cos x}$ Substitute in zero. Answer is $\boxed{\sqrt{2}}$

we can easily get our answer using a simple trigonometry identity

$\Rightarrow L = \displaystyle\lim_{x\rightarrow 0^+} \dfrac{x}{\sqrt{1 - \cos x}}$

Put $1 - \cos(x) = 2\sin^2 \dfrac{x}{2}$

$\Rightarrow L =\displaystyle\lim_{x\rightarrow 0^+} \dfrac{x}{\sqrt{2\sin^2 \dfrac{x}{2}}}$

$When \quad x \rightarrow 0 \Rightarrow \sin x \rightarrow x \quad \quad (\text{As x is very small})$

$\Rightarrow L = \displaystyle\lim_{x\rightarrow 0^+} \dfrac{x}{\sqrt{\dfrac{2 x^2}{4}}}$

$\Rightarrow L = \displaystyle\lim_{x\rightarrow 0^+} \sqrt2 \dfrac{x}{\sqrt{x^2}} \quad, \quad \sqrt{x^2} = x \quad \quad (\text{As x is positive} )$

$\Rightarrow \boxed{L = \sqrt2}$

$\begin{aligned} L & = \lim_{x \to 0^+} \frac x{\sqrt{1-\color{#3D99F6} \cos x}} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0^+} \frac x{\sqrt{1-\color{#3D99F6}1 + \frac {x^2}{2!} - \frac {x^4}{4!} + \cdots}} & \small \color{#3D99F6} \text{Divide up and down by }x \\ & = \lim_{x \to 0^+} \frac 1{\sqrt{\frac 1{2!} - \frac {x^2}{4!} + \frac {x^4}{6!} - \cdots}} \\ & = \sqrt 2 \end{aligned}$

Therefore, $a= \boxed 2$ .