Evaluating $\lim\limits_{x\to0}\frac{1-\cos(x)}{x}$

Question

$$\lim_{x\to0}\frac{1-\cos(x)}{x}$$

Could someone help me with this trigonometric limit? I am trying to evaluate it without l'Hôpital's rule and derivation.

More solutions to this limit are here and here. – Lord_Farin Jun 15 '13 at 07:55 — Lord_Farin, Jun 15 '13 at 07:55

score 12 · Answer 1 · answered Jun 15 '13 at 01:14

12

Hint: Multiply top and bottom by $1+\cos x$. Then use a probably known fact about $\dfrac{\sin x}{x}$.

answered Jun 15 '13 at 01:14

André Nicolas

507,029

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Muito obrigada! – user73276 Jun 15 '13 at 01:16
but when i replace the x per 0? – user73276 Jun 15 '13 at 01:25
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Our expression is equal to $\frac{1}{1+\cos x}\frac{\sin x}{x}\sin x$. As $x\to 0$, the $\frac{\sin x}{x}$ part approaches $1$, and the dangling $\sin x$ part approaches $0$, so the product approaches $0$. – André Nicolas Jun 15 '13 at 01:32

score 9 · Accepted Answer · answered Jun 15 '13 at 01:29

9

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

O you simply apply L'Hopital rule and $$\displaystyle \lim_{x\to 0} \frac{1-\cos x}{x}=\displaystyle \lim_{x\to 0} \frac{(1-\cos x)'}{x'}=\displaystyle \lim_{x\to 0} \frac{\sin x}{1}=0$$

answered Jun 15 '13 at 01:29

math_man

1,574

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or you simply apply L'h – math_man Jun 15 '13 at 01:30
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L'HR makes a lot of trigonometric limit problems super easy, although this particular limit is often introduced in calculus courses well before the Rule appears (since we generally need this limit to obtain the derivatives of sine and cosine). – colormegone Jun 15 '13 at 01:37

not all wrong · Answer 3 · 2013-06-15T01:34:22.440

5

As an alternative method, $\cos x=1+ O(x^2)$. Then $$O(x^2)/x= O(x)$$

Edit: The notation $f(x)=O(g(x))$ means that $f(x)/g(x)$ is bounded by a constant (in the limit). Put simply, this means $f$ is no bigger than $g$ in the limit.

This answer basically relies on the Taylor expansion of the numerator. Typically this is the quickest way to get to the answer with these sorts of limits.

For example, l'Hôpital's rule is seen to be because if $f(0)=g(0)=0$ for functions which can be approximated to some order by a Taylor series then $f(x)/g(x)= (x f'(0)+O(x^2))/(xg'(0)+O(x^2))$ and one can cancel the $x$s.

edited Jun 15 '13 at 01:34

answered Jun 15 '13 at 01:23

not all wrong

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I would suspect (though I don't know for a fact) that anybody asking this type of question would be unfamiliar with $O$-notation - and, assuming this comes from a classroom context, they are probably not allowed these sorts of arguments in any case. – Nick Peterson Jun 15 '13 at 01:24
I've added a quick definition. However, I see the point in answering this type of question to be just providing other ways (possibly more general) to look at the problem. Most mathematicians, confronted with this, do not think "Aha, let us do a clever manipulation to relate it to some standard limit", they think "Errr, that's trivial by Taylor expansion". I see no harm on introducing the idea. – not all wrong Jun 15 '13 at 01:30
Oh, don't get me wrong. I certainly wasn't saying there was any harm. Please, take the comment as coming from a mathematician who has seen WAY too many students try to reach into bags of tricks that they don't understand! – Nick Peterson Jun 15 '13 at 01:32
Fair enough, I suppose! I'm mainly reacting to the "Find a fancy way to do it" school of answer, which is fine, but is worth balancing out. (: – not all wrong Jun 15 '13 at 01:37

score 4 · Answer 4 · answered Jun 15 '13 at 04:44

$$\lim_{x\to0}\frac{1-\cos(x)}{x}$$ $$\lim_{x\to0}\frac{1-(1-2\sin^2 \dfrac x2)}{x}$$ $$\lim_{x\to0}\frac{2\sin^2\dfrac x2}{x}$$ $$\lim_{x\to0}\frac{2x\sin^2\dfrac x2}{4\left(\dfrac x2\right)^2}$$ $$\lim_{x\to0}\frac{x}{2}\cdot \lim_{x\to0}\left(\dfrac{\sin \dfrac x2}{\dfrac x2}\right)^2$$ $$0\cdot 1$$ $$0$$

score 3 · Answer 5 · answered Jun 15 '13 at 07:24

3

Using the definition of derivative: $$\lim\limits_{x \to 0} \frac{1-\cos(x)}{x}= - \lim\limits_{x \to 0} \frac{\cos(x)-\cos(0)}{x-0}=- \cos'(0)=-\sin(0)=0$$

answered Jun 15 '13 at 07:24

Seirios

33,157

score 2 · Answer 6 · answered Jun 15 '13 at 01:25

Write $$\frac{1-\cos x}{x} = \frac{\cos 0 -\cos x }{x} = \frac{-2\sin\left( {0 + x \over 2}\right) \sin\left({0 -x \over 2}\right)}{x} = \frac{\sin^2(x/2)}{x/2}$$ and then
$$ \lim_{x\to 0}\frac{1-\cos x}{x} = \lim_{x\to 0}\frac{\sin^2(x/2)}{x/2} = \lim_{x\to 0} \frac{\sin(x/2)}{x/2}\cdot \lim_{x\to 0}\sin (x/2) . $$

colormegone · Answer 7 · 2013-06-15T03:32:10.220

There is also a fairly direct method based on trig identities and the limit $ \ \lim_{\theta \rightarrow 0} \ \tan \frac{\theta}{2} \ = \ 0 , $ which I discuss in the first half of my post here.

[In brief,

$$ \lim_{\theta \rightarrow 0} \ \tan \frac{\theta}{2} \ = \ 0 \ = \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\sin \theta} \ = \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta} \ \cdot \ \frac{\theta}{\sin \theta} $$

$$= \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta} \ \cdot \ \frac{1}{\lim_{\theta \rightarrow 0} \ \frac{\sin \theta}{\theta} } \ = \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta} \ \cdot \ \frac{1}{1} \ \Rightarrow \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta} \ = \ 0 \ . ] $$

Evaluating $\lim\limits_{x\to0}\frac{1-\cos(x)}{x}$

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