Show that $\lim_{x\to 0}\frac{x(1+\cos (x))-2\tan (x)}{2x-\sin (x)-\tan (x)}=7$

Question

Show that $\lim\limits_{x\to 0}\dfrac{x\left(1+\cos \left(x\right)\right)-2\tan \left(x\right)}{2x-\sin \left(x\right)-\tan \left(x\right)}=7$ ?

Answer 1

The denominator seems simpler; the degree $1$ expansions of sine and tangent are $$ \sin x=x+o(x),\qquad \tan x=x+o(x) $$ so $2x-\sin x-\tan x=o(x)$, which doesn't help. So let's consider the expansion to degree $3$ (they are odd functions, so the degree $2$ term is zero): $$ \sin x=x-\frac{x^3}{6}+o(x^3),\qquad \tan x=x+\frac{x^3}{3}+o(x^3) $$ so the denominator is $$ 2x-\sin x-\tan x=-\frac{x^3}{6}+o(x^3) $$ Now expand the numerator

$$x(1+\cos x)-2\tan x=x\left(1+1-\frac{x^2}{2}+o(x^2)\right)-2x-\frac{2x^3}{3}+o(x^3)=-\frac{7}{6}x^3+o(x^3)$$

Answer 2

By L'Hôpital's rule we obtain: $$\lim\limits_{x\to 0}\dfrac{x\left(1+\cos{x}\right)-2\tan{x}}{2x-\sin {x}-\tan {x}}=\lim_{x\rightarrow0}\frac{1+\cos{x}-x\sin{x}-\frac{2}{\cos^2x}}{2-\cos{x}-\frac{1}{\cos^2x}}=$$ $$\lim_{x\rightarrow0}\frac{\cos^2x+\cos^3x-x\cos^2x\sin{x}-2}{2\cos^2x-\cos^3x-1}=$$ $$=\lim_{x\rightarrow0}\left(\frac{\cos^2x+2\cos{x}+2}{1+\cos{x}-\cos^2x}-\frac{x\cos^2x\sin{x}}{(\cos{x}-1)(1+\cos{x}-\cos^2x)}\right)=$$ $$=\lim_{x\rightarrow0}\left(5-\frac{x\sin{x}}{\cos{x}-1}\right)=\lim_{x\rightarrow0}\left(5+\frac{x\sin{x}}{2\sin^2\frac{x}{2}}\right)=\lim_{x\rightarrow0}\left(5+2\cdot\frac{\frac{x}{2}\cos{\frac{x}{2}}}{\sin\frac{x}{2}}\right)=7.$$