how to do- $$\lim_{x \rightarrow 0} (\cos 2x -1)/(\cos x- 1 )$$

Question

how to do- $$\lim_{x \rightarrow 0} (\cos 2x -1)/(\cos  x- 1 )$$

kira_yamato · Answer

Use l Hospital theorem (I don't know how to spell it) But it goes $$\lim_{x \rightarrow a} \left[ \frac{f(x)}{g(x)} \right] = \lim_{x \rightarrow a} \left[ \frac{f'(x)}{g'(x)} \right]$$

zarkon · Answer

$$\lim_{x \to0} \frac{\cos (2x) -1}{\cos (x)- 1}$$
$$=\lim_{x \to0} \frac{\cos^2(x)-\sin^2(x) -1}{\cos  (x)- 1}$$
$$=\lim_{x \to0} \frac{-(1-\cos^2(x))-\sin^2(x) }{\cos  (x)- 1}$$
$$=\lim_{x \to0} \frac{-\sin^2(x)-\sin^2(x) }{\cos  (x)- 1}$$
$$=\lim_{x \to0} \frac{-2\sin^2(x)}{\cos  (x)- 1}$$
$$=\lim_{x \to0} \frac{-2\sin^2(x)}{\cos  (x)- 1}\frac{\cos  (x)+ 1}{\cos  (x)+ 1}$$
$$=\lim_{x \to0} \frac{-2\sin^2(x)}{\cos^2  (x)- 1}(\cos (x)+ 1)$$
$$=\lim_{x \to0} \frac{-2\sin^2(x)}{-\sin^2 (x)}(\cos  (x)+ 1)$$
$$=\lim_{x \to0}\,\,\,\, 2(\cos  (x)+ 1)=2(1+1)=4$$