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Please help me to solve this limit without using L'Hôpital's rule. I don't know what other method can't be used to solve this limit.

$$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $$

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    $\begingroup$ A start: Multiply top and bottom by $1+\cos x$. $\endgroup$ – André Nicolas Mar 25 '16 at 16:06
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    $\begingroup$ Usually $\lim_{x\to 0}\frac{1-\cos x}x = 0$ is proven at the same time as $\lim_{x\to 0} \frac{\sin x}x = 1$ is, in the development of the derivatives of $\sin$ and $\cos$. Since $\lim_{x\to 0} \frac 1{\cos x} = 1$ by continuity, your limit is $0\cdot 1 = 0$. $\endgroup$ – Paul Sinclair Mar 25 '16 at 16:13
  • $\begingroup$ The answers given here might be useful, too: Evaluating $\lim_{x\to0}\frac{1-\cos(x)}{x}$ $\endgroup$ – Martin Sleziak Apr 28 '18 at 17:57
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HINT:

$$\frac{1-\cos(x)}{x\cos(x)}=\frac{2\sin^2(x/2)}{x\cos(x)} \tag 1$$


SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Using $(1)$ we have $$\begin{align}\lim_{x\to 0}\frac{1-\cos(x)}{x\cos(x)}&=\lim_{x\to 0}\frac{2\sin^2(x/2)}{x\cos(x)}\\\\&=\left(\lim_{x\to 0}\frac{\sin(x/2)}{x/2}\right)\left(\lim_{x\to 0}\frac{\sin(x/2)}{\cos(x)}\right)\\\\&=(1)(0)\\\\&=0\end{align}$$

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  • $\begingroup$ Nice! As usual, I assumed the OP knows Taylor Series lol. This is more grounded, may fit better with his knowledge. $\endgroup$ – Von Neumann Mar 25 '16 at 16:13
  • $\begingroup$ @1over137 Thank you very much. Very appreciative! And you are probably correct. -Mark $\endgroup$ – Mark Viola Mar 25 '16 at 16:16
  • $\begingroup$ +1 I found this answer very useful, not sure how OP feels. $\endgroup$ – John Joy Mar 25 '16 at 16:38
  • $\begingroup$ @johnjoy pleased to hear! And much appreciative of the comment. -Mark $\endgroup$ – Mark Viola Mar 25 '16 at 17:15
  • $\begingroup$ Don't you need l'hopitals for $/frac{sin(x)}{x}$ $\endgroup$ – mike van der naald Mar 25 '16 at 17:16
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This answer is a follow-up to André Nicolas' and Paul Sinclair's comments

$$\lim_{x\to 0}\frac{1-\cos x}{x\cos x}=\lim_{x\to 0}\frac{1-\cos x}{x\cos x}\cdot\frac{1+\cos x}{1+\cos x}=\lim_{x\to 0}\frac{1-\cos^2 x}{x\cos x(1+\cos x)}=\lim_{x\to 0}\frac{\sin^2 x}{x\cos x(1+\cos x)}$$

Note that $\lim_{x\to 0}\frac{\sin x}{x}=1$, that $\lim_{x\to 0}\sin x=0$, that $\lim_{x\to 0}\cos x=1$, and that $\lim_{x\to 0}(1+\cos x)=2$ giving the final result $$\lim_{x\to 0}\frac{1-\cos x}{x\cos x}=0$$

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Use Taylor series

$$\cos(x) \approx 1 - \frac{x^2}{2}$$ thence

$$\frac{1 - \cos(x)}{x\cos(x)} = \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x\cdot\left(1 - \frac{x^2}{2}\right)} = \frac{x^2}{2x - x^3} = \frac{x^2}{x^2\left(\frac{2}{x} - x\right)} = \frac{1}{\frac{2}{x}} = \frac{x}{2}$$

And since $x\to 0$ the limit is

$$\boxed{0}$$

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  • $\begingroup$ +1 Although there are those on this site who attach some equivalence with LHR and Taylor series. $\endgroup$ – Mark Viola Mar 25 '16 at 17:35
  • $\begingroup$ @Dr.MV Super thank you! Maybe Taylor Series can be inappropriate sometimes, but I thought in that case it held ^^ $\endgroup$ – Von Neumann Mar 25 '16 at 17:46
  • $\begingroup$ Yes, absolutely. I actually prefer asymptotics over LHR for many situations. $\endgroup$ – Mark Viola Mar 25 '16 at 18:47
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Note that $ \displaystyle \lim_{x\to0}\dfrac{1-\cos x}{x} =-\lim_{x\to0}\dfrac{\cos x-\cos 0}{x - 0} =-\left.\cos'x\right|_{x=0} = \sin 0 = 0 $

So $ \displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} = \lim_{x\to 0}\frac{1-\cos(x)}{x} \cdot \lim_{x\to 0}\dfrac{1}{\cos(x)} = 0 \cdot 1 = 0 $

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  • $\begingroup$ But how do you prove $cos'(x) = - sin(x)$? $\endgroup$ – user261263 Mar 25 '16 at 17:03
  • $\begingroup$ @EugenCovaci - If you're at the point where you know what L'Hôpital's rule is, you should already know. $\endgroup$ – steven gregory Mar 25 '16 at 17:08
  • $\begingroup$ The question is: how do you prove $cos'(x) = - sin(x)$ without L'Hôpital's rule? $\endgroup$ – user261263 Mar 25 '16 at 17:12
  • $\begingroup$ @EugenCovaci - where does it say that? $\endgroup$ – steven gregory Mar 25 '16 at 17:18
  • $\begingroup$ "Please help me to solve this limit without using L'Hôpital's rule" I suppose this means avoinding direct or indirect use of L'Hôpital's rule. $\endgroup$ – user261263 Mar 25 '16 at 17:19
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$\frac{1-\cos x}{x\cos x} = \frac{2 \sin^2 (x/2)}{x \cos x}= \frac{\sin (x/2)}{x/2} \frac{\sin (x/2)}{\cos x}$ The limit is $0$.

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  • $\begingroup$ Yes! And what does this mean? $\endgroup$ – kmitov Mar 25 '16 at 16:13
  • $\begingroup$ It means that you write faster than me. Nothing els, sir. $\endgroup$ – kmitov Mar 25 '16 at 16:16
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frist :

$$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{2}$$

now :

$$ \lim_{x \to 0} \frac{1-\cos x}{x\cos x}=\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x^2}{x\cos x}\\=\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x}{\cos x}=?$$

since:

$$\lim_{x \to 0} \frac{x}{\cos x}=0$$

so :

$$\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x}{\cos x}=(\frac{1}{2})(0)=0$$

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