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$$\lim_{x \to 0}\frac{\cos 2x-1}{\cos x-1}$$ I have found the above limit using L'Hopital's rule but since this rule is not given in the book so I'm supposed to do it without using this rule.

I know $$\lim_{x \to 0}\frac{1-\cos x}{x}=0$$

I tried to get something of the form of the above limit but I failed to do so.

Kindly help me solve this problem without using L'Hopital's rule.

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  • $\begingroup$ Have you tried using trigonometric identities to make it less daunting? I believe there is one that will help. $\endgroup$
    – Terra Hyde
    Jul 15, 2015 at 3:17
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    $\begingroup$ @TerraHyde Sir are you saying to use $cos(2x)=2cos^2(x)-1$. $\endgroup$
    – Singh
    Jul 15, 2015 at 3:19
  • $\begingroup$ In particular, you can use a trig identity to get rid of the 2 inside the cosine in the numerator! $\endgroup$ Jul 15, 2015 at 3:19
  • $\begingroup$ @Singh Yes, I was suggesting exactly that. Then it becomes very readily solvable. $\endgroup$
    – Terra Hyde
    Jul 15, 2015 at 3:23

5 Answers 5

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Recall $\cos(2x)=2\cos^2(x)-1$ so we may rewrite as

$$\lim\limits_{x\to 0} 2\frac{\cos^2(x)-1}{\cos(x)-1}=\lim\limits_{x\to 0} 2\frac{(\cos(x)-1)(\cos(x)+1)}{\cos(x)-1}=2(\cos(0)+1)=4$$

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Since others have already presented a great way forward, I thought that it would be instructive to present an alternative approach. To that end, we use

$$\cos x=1-\frac12 x^2+O(x^4)$$

to write

$$\begin{align}\frac{\cos 2x-1}{\cos x-1}&=\frac{\left(1-\frac12 (2x)^2+O(x^4)\right)-1}{\left(1-\frac12 x^2+O(x^4)\right)-1}\\\\&=\frac{-2x^2+O(x^4)}{-\frac12 x^2+O(x^4)}\\\\&=4+O(x^4)\to 4\end{align}$$

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    $\begingroup$ this is equivalent to using l'Hôpital... $\endgroup$
    – hjhjhj57
    Jul 15, 2015 at 5:14
  • $\begingroup$ @hjhjhj57 Well, it isn't really equivalent. We don't have the appearance of $\frac{-2\sin 2x}{-\sin x}$ herein, which by the way would require a second round of LR to get to $\frac{-4\cos 2x}{-\cos x}$. Asymptotic expansions are not equivalent. $\endgroup$
    – Mark Viola
    Jul 15, 2015 at 5:28
  • $\begingroup$ It's equivalent in the sense that both methods require the use of derivatives, in fact, you can see l'Hôpital as a simple corollary of this approach. $\endgroup$
    – hjhjhj57
    Jul 15, 2015 at 5:49
  • $\begingroup$ Not quite. First, if the cosine function is DEFINED in terms of its Taylor series, then we never formally took a derivative. And there are other ways to obtain the bounds used herein. On the second point ... not quite... the proof of LR typically relies on the extended MVT for which $\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\xi)}{g'(\xi)}$ under the usual hypothesis. ;-)) $\endgroup$
    – Mark Viola
    Jul 15, 2015 at 6:08
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$\dfrac{1-\cos (2x)}{1-\cos x} = \dfrac{2\sin^2x}{2\sin^2\left(\frac{x}{2}\right)}=4\cdot \left(\dfrac{\sin x}{x}\right)^2\cdot \left(\dfrac{\dfrac{x}{2}}{\sin\left(\dfrac{x}{2}\right)}\right)^2\to 4\cdot 1\cdot 1=4$, as $x \to 0$.

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Hint: try to expand $\cos 2x$ first.

Remember $\cos 2x=\cos^2x-\sin^2x$, or $\cos(2x)=2\cos^2x-1$. You can continue solve this problem by using this identity.

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knowing that : when x=0 \begin{align*} 1-cos(f(x))\sim \dfrac{1}{2} {f(x)}^{2} \newline \newline \lim_{x \rightarrow 0} \dfrac{cos2x-1}{cosx-1} = \lim_{x \rightarrow 0} \dfrac{-(-cos2x+1)}{-(-cosx+1)} = \lim_{x \rightarrow 0} \dfrac{\dfrac{1}{2}{(2x)}^2}{\dfrac{1}{2}{x}^2}= \lim_{x \rightarrow 0} \dfrac{{(2x)}^2}{{x}^2} = \lim_{x \rightarrow 0} \dfrac{4{x}^2}{{x}^2} = 4 \end{align*}

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