## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{x-\sin x}{x-\tan x}=-\frac{1}{2}$$
$$A=\lim_{x\to0}\frac{x-\sin x}{x-\tan x}$$ As $x\to0$, $(x-\sin x)\to(0-\sin0)=0-0=0$ and $(x-\tan x)\to(0-\tan0)=0-0=0.$ And we have an indeterminate form of $\frac{0}{0}$, ready for the application of L'Hospital's Rule: $$A=\lim_{x\to0}\frac{(x-\sin x)'}{(x-\tan x)'}$$ $$A=\lim_{x\to0}\frac{1-\cos x}{1-\sec^2 x}$$ Now you can do L'Hospital's Rule one more time since this is an indeterminate form of $\frac{0}{0}$. However, since we've eliminated the $x$ elements, we can apply trigonometric rules to simplify the formula, which is the way I would do here. $$A=\lim_{x\to0}\frac{1-\cos x}{1-\frac{1}{\cos^2 x}}$$ $$A=\lim_{x\to0}\frac{1-\cos x}{\frac{\cos^2 x-1}{\cos^2 x}}$$ $$A=\lim_{x\to0}\frac{\cos^2 x(1-\cos x)}{\cos^2x-1}$$ $$A=\lim_{x\to0}\frac{-\cos^2 x(\cos x-1)}{(\cos x-1)(\cos x+1)}$$ $$A=\lim_{x\to0}\frac{-\cos^2 x}{\cos x+1}$$ Now we can do the replacement: $$A=\frac{-\cos^20}{\cos0+1}$$ $$A=\frac{-1^2}{1+1}$$ $$A=-\frac{1}{2}$$