Find $\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$

Question

I'm trying to find the following limit:

$$\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$$

I tried to use squeeze theorem but it's not making much sense. I did the following:

$$\begin{align} &\lim_{x\rightarrow 0} \frac{\cos x - 1}{x} \\ -1 \le &\lim_{x\rightarrow 0} (\cos x)(x^{-1}) \le 1 \\ \lim_{x\rightarrow 0} -x^{-1} \le &\lim_{x\rightarrow 0} \cos x \le \lim_{x\rightarrow 0} x^{-1} \quad \text{*} \end{align}$$

The last line is where I'm confused. I don't think I'm doing squeeze theorem correctly. I'm guessing you have to manipulate $\cos x - 1$ somehow. Please provide some hints.

Thanks a bunch!

P.S You cannot use L'Hopital.

Hint: $$\lim_{x \to 0} \frac{\cos(x) - 1}{x} = \lim_{x \to 0} \frac{\cos(x) - \cos(0)}{x - 0}.$$ — Tom, Nov 04 '13 at 20:38
Why can't we use L'Hospital? (Not that I'd want to.) You can use a Taylor approximation of the cosine. — Daniel Fischer, Nov 04 '13 at 20:38
I'm aware of both L'Hospital and Taylor aprrox but we haven't covered it in class so I can't really use that. — Jeel Shah, Nov 04 '13 at 20:40
The tag ([tag:limit-theorems]) is intended for questions about limit theorems in probability theory and not for questions about determining limits of sequences or functions, see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) — Martin Sleziak, Nov 29 '13 at 12:57
possible duplicate of Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem — , Sep 02 '14 at 00:44
@JLA There are proofs for the derivative of cosine not relying on either the limit or defining cosine by its Maclaurin series, though. — Divide1918, Jan 26 '24 at 17:47
@JLA If you use a proof for derivative of cosine not relying on the limit, how is using the derivative of cosine (e.g. via L'Hopital's rule) to solve the limit circular? — Divide1918, Jan 28 '24 at 08:02
@Divide1918 The question is asking to find the derivative of $\cos{x}$ at $0,$ using Hopital's rule requires prior knowledge of what the derivative is. — JLA, Jan 28 '24 at 17:27
@JLA I believe you ought to read my first reply again. I reckon you didn’t quite comprehend it. — Divide1918, Jan 29 '24 at 03:33
@Divide1918 Does your definition of the derivative also not involve a limit? If not, then you're right I don't understand what your argument is. Perhaps you should just write it out. — JLA, Jan 29 '24 at 07:17

score 3 · Answer 1 · answered Nov 04 '13 at 20:39

3

Hint 1: $$\cos x -1 = -2\sin^2 \dfrac{x}2$$ Hint 2: $$\dfrac{\cos x - 1}{x} = \dfrac{(\cos x - 1)(\cos x + 1)}{x(\cos x + 1)} = -\dfrac{\sin^2 x}{x^2} \dfrac{x}{1+\cos x}.$$

answered Nov 04 '13 at 20:39

njguliyev

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I tried the second method prior to squeeze but it didn't work. I suspect, I didn't simplify properly. Once you expand the top, you get $\sin^2x - 1$ then do you convert $\sin^2x$ into something or do you convert the $1$ into something? – Jeel Shah Nov 04 '13 at 20:43
We get $\cos^2 x-1 = -\sin^2 x$, not $\sin^2 x -1$. – njguliyev Nov 04 '13 at 20:46
Sorry, does the $\frac{-\sin^2x}{x^2}$ become $\frac{x}{1+\cos x}$ or is it all one thing being multiplied? – Jeel Shah Nov 04 '13 at 20:51
They are being multiplied. I wrote it in this form because we know that $\dfrac{\sin x}{x} \to 1$, as $x \to 0$. – njguliyev Nov 04 '13 at 20:53
So in essence what is happening is that the $\sin x^2x$ fraction is going to $-1$ while the other is going $0$ therefore, the limit is zero right? – Jeel Shah Nov 04 '13 at 21:21
Yes. ${}{}{}{}{}$ – njguliyev Nov 04 '13 at 21:23
let us continue this discussion in chat – Jeel Shah Nov 04 '13 at 21:26

Barry Cipra · Answer 2 · 2013-11-04T21:31:39.097

If you know that $\sin \theta \lt \theta$ when $\theta\gt0$, then you can argue that

$$0\lt\left| {1-\cos x\over x}\right|\lt\left|{1-\cos^2x\over x}\right|=\left|{\sin^2x\over x}\right|\lt \left|\sin x\right|$$

at which point the Squeeze Theorem takes over. If you need to prove the inequality $\sin\theta\lt\theta$, note that ${1\over2}\sin\theta$ (for small $\theta$, at least) is the area of the triangle inside the unit circle with vertices at the origin, $(1,0)$, and $(\cos\theta,\sin\theta)$, whereas ${1\over2}\theta$ is the area of the sector containing the triangle. (Alternatively, note that $\sin\theta$, being the vertical distance from $(\cos\theta,\sin\theta)$ to the $x$-axis, is less than the distance from $(\cos\theta,\sin\theta)$ to $(1,0)$, which is less than the arc length connecting those two points, namely $\theta$.)

score 2 · Answer 3 · answered Nov 04 '13 at 21:28

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

By Mean Value Theorem: $$ \verts{\cos\pars{x} - 1 \over x} = \verts{\sin\pars{\xi}} \quad\mbox{where}\quad \left\vert% \begin{array}{lcl} \quad 0 < \xi < x & \mbox{if} & x > 0 \\ \mbox{or}&& \\ \quad x < \xi < 0 & \mbox{if} & x < 0 \end{array}\right. $$

Now, use 'Sandwich Theorem'.

$\large\tt \mbox{Mean Value Theorem}\ --->\quad$ http://en.wikipedia.org/wiki/Mean_value_theorem — Felix Marin, Nov 04 '13 at 21:35
It's been a long time since I took calculus, but don't the differentiation formulas for sine and cosine come after the limits of $\frac{\sin x}x$ and $\frac{1-\cos x}x$ as $x\to0$? — bof, Nov 04 '13 at 22:02
That's $0$k. Any way it's time to refresh it. Anybody over here is refreshing somehow every day. Don't worry about. — Felix Marin, Nov 04 '13 at 22:06

score 0 · Answer 4 · answered Jan 26 '24 at 16:21

0

As an alternative

$$\frac{\cos x - 1}{x}=\frac {\sin x}x\frac{\cos x - 1}{\sin x}$$

with $\frac {\sin x}x\to 1 $ and by tangent half-angle or double angle identities $$\frac{\cos x - 1}{\sin x}=-\tan \frac x 2 \to 0$$

answered Jan 26 '24 at 16:21

user

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score 0 · Answer 5 · answered Nov 04 '13 at 21:22

0

Let $f(x) = \cos(x)$, then $\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = f'(0) = -\sin(0) = 0$

answered Nov 04 '13 at 21:22

Arthur

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The limit the OP asked about is used to prove the differentiation formula for $\cos x$. It would be circular to use the derivative of $\cos x$ to prove the limit. – bof Nov 04 '13 at 22:06
@bof There are proofs for the derivative of cosine not relying on either the limit or defining cosine by its Maclaurin series, though. – Divide1918 Jan 26 '24 at 17:50

score -1 · Answer 6 · edited Nov 29 '13 at 13:46

-1

It is sufficient to use the asymptotic development of $\cos x$ around $0$, namely

$$\cos x = 1 - \frac{x^2}{2} + o(x^3).$$

edited Nov 29 '13 at 13:46

Lord_Farin

17,743

answered Nov 29 '13 at 13:26

Rodney Coleman

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MacLaurin/Taylor series are not available, because this limit is used to derive them -- circularity lurks. – Lord_Farin Nov 29 '13 at 13:50
Maybe I wasn't clear enough. From the a.d. we obtain cos x-1 = -x^2/2 + o(x^3). If we divide by x, for x small and non-zero, then we obtain (cos x-1)/x = -x/2 + o(x^2). Now letting x go tO 0 we obtain the limit 0. – Rodney Coleman Nov 30 '13 at 15:58
I made a typing error: go to 0. – Rodney Coleman Nov 30 '13 at 16:00
@Lord_Farin There are proofs for the derivative of cosine not relying on either the limit or defining cosine by its Maclaurin series. And of course, if you define cosine by its Maclaurin series, using it wouldn’t be circular. – Divide1918 Jan 26 '24 at 17:48

Find $\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$

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