# Solve this limit (Maclaurin or differentiate?)

I have this assignment where I should calculate the limit below: $$\lim_{x\to0}\frac{\sin 2x}{x\cos x}$$ I can use l'Hospitals rule (because it is a "zero divided by zero"-case) and therefore differentiate: $$f(x) = \lim_{x\to0}\frac{\sin 2x}{x\cos x} = \lim_{x\to0}\frac{2\sin x\cos x}{x\cos x}$$ $$f'(x) = \lim_{x\to0}\frac{2(\cos^2 x - \sin^2 x)}{-\sin x}$$ I don't know if this is the right way to go, if there is, I need to extract this more because $\sin(0)=0$.

Option number two is to use Maclaurin: $$\lim_{x\to0}\frac{2\sin x\cos x}{x\cos x} = \lim_{x\to0}\frac{2(x-\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7))(1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6))}{x(1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6))}$$ Am I on the right way in some case above? And If I am, how do I handle O-notations?

• $\sin (2x) = 2\sin x\cos x$. – Daniel Fischer Jan 27 '14 at 12:58
• If you already had $\;\frac{2\sin x\cos x}{x\cos x}\;$ , why in the world didn't you cancel the cosines?? And after that, the derivative is wrong... – DonAntonio Jan 27 '14 at 13:00
• I don't understand why would anywone downvote this question: the OP is showing some self work! He may be wrong, but that's not what downvotes are for, imo. – DonAntonio Jan 27 '14 at 13:07
• @DonAntonio, I don't know, that solution was too easy! – theva Jan 27 '14 at 13:09
• @DonAntonio Write your comment as an answer so that I can mark it as accepted! – theva Jan 27 '14 at 13:13

$$\frac{\sin 2x}{x\cos x}=\frac{\sin 2x}{2x}\cdot\frac2{\cos x}\xrightarrow[x\to 0]{}\;\ldots$$
Also, you can use the equivalence $$\lim_{x\to 0}{\sin 2x\over 2x}=1.$$
• Is $x\cos x = 2x$? – theva Jan 27 '14 at 13:12
• Isn't. Multiply and divide by $2x$ and use the equivalence. – Martín-Blas Pérez Pinilla Jan 27 '14 at 13:15