0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ $\sin (2x) = 2\sin x\cos x$. $\endgroup$ – Daniel Fischer Jan 27 '14 at 12:58
  • $\begingroup$ 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... $\endgroup$ – DonAntonio Jan 27 '14 at 13:00
  • 1
    $\begingroup$ 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. $\endgroup$ – DonAntonio Jan 27 '14 at 13:07
  • $\begingroup$ @DonAntonio, I don't know, that solution was too easy! $\endgroup$ – theva Jan 27 '14 at 13:09
  • $\begingroup$ @DonAntonio Write your comment as an answer so that I can mark it as accepted! $\endgroup$ – theva Jan 27 '14 at 13:13
4
$\begingroup$

$$\frac{\sin 2x}{x\cos x}=\frac{\sin 2x}{2x}\cdot\frac2{\cos x}\xrightarrow[x\to 0]{}\;\ldots$$

$\endgroup$
0
$\begingroup$

Also, you can use the equivalence $$\lim_{x\to 0}{\sin 2x\over 2x}=1.$$

$\endgroup$
  • $\begingroup$ Is $ x\cos x = 2x $? $\endgroup$ – theva Jan 27 '14 at 13:12
  • 2
    $\begingroup$ Isn't. Multiply and divide by $2x$ and use the equivalence. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 27 '14 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.