Question: Isn't $\lim\_{x \to 0} \frac{\sin 2x^{\circ}}{x^{\circ}}=2$?
But the actual answer doesn't seem to have this rather answers of the form of radians.Where did i go wrong?

Answer: 
For an angle of $a$ radians and $x$ degrees it is $\dfrac{a}{\pi}=\dfrac{x}{180}\Rightarrow x=\dfrac{180a}{\pi} $ or $a=\dfrac{\pi x}{180}$

In your limit set $a=\dfrac{\pi x}{180}$. Since $x\to 0$ it will be $a\to 0$ and therefore

$\lim\limits\_{x\to 0^{\circ}}\dfrac{sin(2x^{\circ})}{x^{\circ}}=\lim\limits\_{a\to 0}\dfrac{sin(2\frac{180a}{\pi})}{\frac{180a}{\pi}}=\dfrac{2\frac{180}{\pi}}{\frac{180}{\pi}}=2$
Since you can convert degrees to radians, by use of the formula $a\_r = \frac{\pi}{180} \cdot a\_d$, the expression becomes:

$\lim\_{x \rightarrow 0} \frac{2 \sin(u) \cos(u)}{u}$, where $u = \frac{x\pi}{180}$, so clearly this is also equal to $\lim\_{u \rightarrow 0} \frac{2\sin(u) \cos(u)}{u}$. Using the fact that $\lim\_{u \rightarrow 0} \frac{\sin(u)}{u} = 1$, the expression becomes $\lim\_{u \rightarrow 0} 2\cos(u)$, which is clearly equal to $2$