Evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$ Calculate the following limit : 
$$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$
This is what I have tried:
Using Maclaurin series for $ (1+x)^a $:
$$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$
Using Maclaurin series for $\cos x $:
$$\cos x=1-\frac{x^2}{2!}\quad \text{(We'll stop at order 2)}$$
This leads to :
$$1-\sqrt{1+x^2}\cos x=1-(1+\frac{x^2}{2})(1-\frac{x^2}{2})=\frac{x^4}{4}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4$$
Using Maclaurin series for $\sin x $:
$$\sin x=x-\frac{x^3}{3!}\quad \text{(We'll stop at order 3)}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4 = \left(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}}\right)^4$$
Thus $$\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\frac{\frac{x^4}{4}}{(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}})^4}=\frac{x^4(1-\frac{x^2}{2})^4}{4(x-\frac{x^3}{3!})}=\frac{(x(1-\frac{x^2}{2}))^4}{4(x-\frac{x^3}{3!})}=\frac{1}{4}(\frac{x-\frac{x^3}{2}}{x-\frac{x^3}{3!}})=\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})
$$
Then $$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\lim_{x\rightarrow 0}\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})=\frac{1}{4}.$$
But my book says the solution is $\frac{1}{3}$
Where have I done wrong?
Help appreciated!

Without using L'Hospital & Taylor's Expansion,
$$=\dfrac{1-(1+x^2)(1-\sin^2x)}{1+\cos x\sqrt{1+x^2}}\cdot\dfrac{\cos^4x}{\sin^4x}$$
$$=\dfrac{x^2\sin^2x+(\sin x-x)(\sin x+x)}{x^4}\cdot\dfrac{\cos^4x}{\left(\dfrac{\sin x}x\right)^4(1+\cos x\sqrt{1+x^2})}$$
Now $\dfrac{(\sin x-x)(\sin x+x)}{x^4}=\left(\dfrac{\sin x}x+1\right)\cdot\dfrac{(\sin x-x)}{x^3}$
Now use Are all limits solvable without L'Hôpital Rule or Series Expansion
