Evaluate the limit $\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$ I need to evaluate the limit without using l'Hopital's rule.
$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$
 A: $\frac{\sqrt{1+x\sin(x)}-\sqrt{\cos(2x)}}{\tan^{2}(\frac{x}{2})}=\bigg(\frac{1-\cos(2x)}{\tan^{2}(\frac{x}{2})}+\frac{x\sin(x)}{\tan^{2}(\frac{x}{2})}\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}=\bigg(\frac{1-\cos(2x)}{(2x)^{2}}\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}(16\cos^{2}(\frac{x}{2}))+\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}\frac{\sin(x)}{x}(4\cos^{2}(\frac{x}{2}))\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}$
Letting $x$ go to $0$ in the last formulation of the limit gives:
$(\frac{1}{2}\cdot1\cdot16+1\cdot1\cdot4)\frac{1}{2}=6$.
A: If we are allowed to use Taylor series: 
Since $\sin x = x + O(x^3)$, we have $1+x\sin x = 1 + x^2 + O(x^4)$. 
Thus, $\sqrt{1+x\sin x} = 1+\dfrac{1}{2}x^2 + O(x^4)$. 
Since $\cos x = 1 - \dfrac{1}{2}x^2 + O(x^4)$, we have $\cos 2x = 1 - 2x^2 + O(x^4)$. 
Thus, $\sqrt{\cos 2x} = 1-x^2 + O(x^4)$. 
Since $\tan x = x + O(x^3)$, we have $\tan \dfrac{x}{2} = \dfrac{1}{2}x + O(x^3)$.
Thus, $\tan^2 \dfrac{x}{2} = \dfrac{1}{4}x^2 + O(x^4)$.
Therefore, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} = \dfrac{\left(1+\tfrac{1}{2}x^2 + O(x^4)\right) - \left(1-x^2+O(x^4)\right)}{\tfrac{1}{4}x^2 + O(x^4)} = \dfrac{\tfrac{3}{2}x^2+O(x^4)}{\tfrac{1}{4}x^2+O(x^4)} = \dfrac{\tfrac{3}{2}+O(x^2)}{\tfrac{1}{4}+O(x^2)}$. 
Hence, as $x \to 0$, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} \to \dfrac{\tfrac{3}{2}}{\tfrac{1}{4}} = 6$. 
