Using Taylor series to evaluate $\lim_{x\to0} \frac{\sqrt{1+ x\sin(x)} - \sqrt{\cos(2x)}}{ \tan^2(x/2)}$ To be honest, I have to solve the following exercise, but I don't know what trigonometric formulas should I use (if I should), to get to a form where I can use the Taylor series.
$$\lim_{x\rightarrow 0} \frac{\sqrt{1+ x\sin(x)} - \sqrt{\cos(2x)}}{ \tan^2(x/2)}$$
I have tried to convert the tan function (using trig formulas) but it didn't work, also I tried to use Taylor series on the $\sin(x), \cos(2x)$ functions.
 A: $$L=\lim_{x\rightarrow 0} \frac{{\sqrt{1+ x\sin(x)}} - \sqrt{\cos(2x)}}{\tan^2(x/2)}$$
$$L=\lim_{x\rightarrow 0} \frac{{1+ x\sin(x)} - \cos(2x)}{\tan^2(x/2)\Bigg({\sqrt{1+ x\sin(x)}} + \sqrt{\cos(2x)}\Bigg)}$$
$$L=\lim_{x\rightarrow 0} \frac{{ x\sin(x)} + 2\sin^2x}{\tan^2(x/2)\Bigg({\sqrt{1+ x\sin(x)}} + \sqrt{\cos(2x)}\Bigg)}$$
$$L=\lim_{x\rightarrow 0} \frac{{ \frac{\sin(x)}{x}} + 2.\frac{\sin^2x}{x^2}}{\frac14.\frac{\tan^2(x/2)}{(x/2)^2}\Bigg({\sqrt{1+ x\sin(x)}} + \sqrt{\cos(2x)}\Bigg)}\ \text{Dividing $N^r$ and $D^r$ by $x^2$}$$
Now use $\lim_{t\to0}\frac{\sin t}{t}=1$ and $\lim_{t\to0}\frac{\tan t}{t}=1$ to obtain
$$\boxed{L=6}$$
A: When $z$ is small,
$\sin z =z+..$, $\cos z=1-z^2/2$, $\tan z=z+...$
Then $$L=\lim_{x\to 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos 2x}}{\tan^2(x/2)}= \lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-2x^2}}{x^2/4}$$
Use binomial approximation that $(1+z)^k=1+kz$ if $z$ is very small, then
$$L=\lim_{x \to 0} \frac{(1+x^2/2)-(1-x^2)}{x^2/4}=6.$$
A: As noticed we don't need Taylor's series in this case but if we want to proceed by this way we have that

*

*$x\sin x=x^2+O(x^3) \implies \sqrt{1+ x\sin(x)}=1+\frac12 x^2+O(x^3)$

*$\cos(2x)=1-2x^2 +O(x^3) \implies \sqrt{\cos(2x)}=1-x^2+O(x^3)$

*$\tan^2(x/2)=\frac14 x^2+O(x^3)$
and therefore
$$\frac{\sqrt{1+ x\sin(x)} - \sqrt{\cos(2x)}}{ \tan^2(x/2)}=$$
$$=\frac{1+\frac12 x^2-1+x^2+O(x^3)}{\frac14 x^2+O(x^3)}=\frac{\frac32 x^2+O(x^3)}{\frac14 x^2+O(x^3)}=\frac{\frac32+O(x)}{\frac14 + O(x)} \to 6$$
A: The arguments of the two square roots will expand as $1$ and a quadratic term ($x^2$ and $-2x^2$ respectively), and the square roots will halve those quadratic terms. The $1$'s cancel out and the quadratic terms end-up as $\dfrac{3x^2}2$.
The denominator is approximated by $\dfrac{x^2}4$, and finally the ratio tends to the finite limit $$6.$$
