TITLE: Evaluate the limit $\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$
QUESTION [5 upvotes]: I need to evaluate the limit without using l'Hopital's rule.
$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$

REPLY [2 votes]: $\frac{\sqrt{1+x\sin(x)}-\sqrt{\cos(2x)}}{\tan^{2}(\frac{x}{2})}=\bigg(\frac{1-\cos(2x)}{\tan^{2}(\frac{x}{2})}+\frac{x\sin(x)}{\tan^{2}(\frac{x}{2})}\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}=\bigg(\frac{1-\cos(2x)}{(2x)^{2}}\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}(16\cos^{2}(\frac{x}{2}))+\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}\frac{\sin(x)}{x}(4\cos^{2}(\frac{x}{2}))\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}$
Letting $x$ go to $0$ in the last formulation of the limit gives:
$(\frac{1}{2}\cdot1\cdot16+1\cdot1\cdot4)\frac{1}{2}=6$.