8
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Have you tried using trigonometric identities to make it less daunting? I believe there is one that will help. $\endgroup$ – Terra Hyde Jul 15 '15 at 3:17
  • 1
    $\begingroup$ @TerraHyde Sir are you saying to use $cos(2x)=2cos^2(x)-1$. $\endgroup$ – Singh Jul 15 '15 at 3:19
  • $\begingroup$ In particular, you can use a trig identity to get rid of the 2 inside the cosine in the numerator! $\endgroup$ – Trevor J Richards Jul 15 '15 at 3:19
  • $\begingroup$ @Singh Yes, I was suggesting exactly that. Then it becomes very readily solvable. $\endgroup$ – Terra Hyde Jul 15 '15 at 3:23
9
$\begingroup$

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

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

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ this is equivalent to using l'Hôpital... $\endgroup$ – hjhjhj57 Jul 15 '15 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 '15 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 '15 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 '15 at 6:08
2
$\begingroup$

Hint: try to expand cos 2x first.

Remember cos 2x = cos² x - sin² x, or cos 2x = 2cos² x - 1. You can continue solve this problem by using this identity.

$\endgroup$
2
$\begingroup$

$\dfrac{1-\cos (2x)}{1-\cos x} = \dfrac{2\sin^2x}{2\sin^2\left(\frac{x}{2}\right)}=4\cdot \left(\dfrac{\sin x}{x}\right)^2\cdot \left(\dfrac{\dfrac{x}{2}}{\sin\left(\dfrac{x}{2}\right)}\right)^2\to 4\cdot 1\cdot 1=4$, as $x \to 0$.

$\endgroup$

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.