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I need help understanding how this limit is proved? :

Show that $$\lim_{h\to 0} \frac{\cos (h)-1}{h}=0$$ Proof:
Using the half angle formula, $\cos h = 1-2 \sin^2(h/2)$

$$\lim_{h\to 0} \frac{\cos (h)-1}{h}\\=\lim_{h\to 0}( -\frac{2 \sin^2(h/2)}{h})\\=-\lim_{\theta \to 0}\frac{\sin \theta}{\theta} \sin \theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Let $\theta=h/2$)} \\ = -(1)(0)\\=0$$ I have no idea how this proof is done, so I apologize for the lack of my own thoughts in this question. I understand limits and know sin, cos, tan, but I am just very lost as what they did in each step. Can someone please explain all the steps of the proof as well as the half-angle formula. Thanks!

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  • $\begingroup$ In the second-to last step you are using $\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$. The rest is just plugging in. $\endgroup$ – Daniel Jul 6 '16 at 21:49
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    $\begingroup$ Another proof is to use the difference quotient : $\lim_{h\to 0} \frac{\cos (h)-1}{h} = \lim_{h\to 0} \frac{\cos (h)-\cos(0)}{h-0} = \cos'(0) = -\sin(0) =0$. $\endgroup$ – anonymus Jul 6 '16 at 21:53
  • $\begingroup$ @anonymus: I'm pretty sure this proof is being done to establish the derivative formulas for $\sin$ and $\cos$. $\endgroup$ – Ted Shifrin Jul 6 '16 at 22:00
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    $\begingroup$ @anonymus: The problem is that, in some derivations, this limit has to be proved to deduce the derivative of $\cos$ is $-\sin$. $\endgroup$ – Bernard Jul 6 '16 at 22:03
  • $\begingroup$ Ow ok, i see. Thanks. $\endgroup$ – anonymus Jul 6 '16 at 22:06
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The simplest proof is this: $$\frac{\cos h-1}h=\frac{(\cos h-1)(\cos h+1)}{(\cos h+1)h}=\frac{\cos^2h-1}{(\cos h+1)h}=-\frac{\sin^2h}{(\cos h+1)h}=-\frac{\sin h}h\cdot\frac{\sin h}{\cos h+1}.$$ The first fraction tends to $1$, the second tends to $\dfrac 02=0$, hence the limit is $\color{red}0$.

For the proof you mention, at the third line, you should have $$=\lim_{h\to 0}\Bigl( -\frac{2 \sin^2(h/2)}{h}\Bigr)=\lim_{h\to 0}\Bigl( -\frac{\sin^2(h/2)}{h/2}\Bigr)=\dots$$

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  • $\begingroup$ I agree: using the half-angle formula and substitution is unnecessary complication, as you have to use the product theorem anyway. $\endgroup$ – egreg Jul 6 '16 at 23:01
  • $\begingroup$ Why is this "the simplest" ? $\endgroup$ – Yves Daoust Nov 25 '20 at 10:49
  • $\begingroup$ @YvesDaoust: I meant the simplest with elementary tools (and to the best of my knowledge). $\endgroup$ – Bernard Nov 25 '20 at 10:56
  • $\begingroup$ But it has as many steps as the OP's version. $\endgroup$ – Yves Daoust Nov 25 '20 at 11:00
  • $\begingroup$ Because I added obvious details, which were unnecessary. $\endgroup$ – Bernard Nov 25 '20 at 11:04
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The first couple of steps just deal with rewriting the expression $$\frac{\cos(h)-1}{h}$$ without taking a limit anywhere. With the formula $\cos(h)=1-2\sin^2\left(h/2\right)$ you can indeed rewrite this to $$-2\frac{\sin^2(h/2)}{h}=-\frac{\sin^2(h/2)}{h/2}=-\frac{\sin^2(\theta)}{\theta}= -\frac{\sin(\theta)}{\theta}\cdot \sin(\theta),$$ for $\theta = h/2$. Now taking the limit $h\to 0$ is the same as taking $\theta \to 0$, so we can now calculate the limit. We can take the minus sign out of the limit, so we have $$ \lim_{\theta \to 0}-\frac{\sin(\theta)}{\theta}\cdot \sin(\theta)=-\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}\cdot \sin(\theta). $$ Now comes the tricky part. If you have two functions $f,g$ for which the limits $\displaystyle\lim_{x->0}f(x)=a$ and $\displaystyle\lim_{x->0}g(x)=b$ both exist, then the limit $\displaystyle \lim_{x->0} f(x)\cdot g(x)$ exists and is equal to $a\cdot b$.

We can apply this theorem to our limit. Concluding we get $$ \lim_{h\to 0} \frac{\cos(h)-1}{h} = -\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}\cdot \sin(\theta) = -\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} \cdot \lim_{\theta \to 0} \sin(\theta) = - 1 \cdot 0 = 0. $$

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There's a way to find the limit $l=\lim\limits_{h\to 0} \frac{cos\ h -1}{h}$ without using the limit of the sinc integral. This is purely algebraic.(Unlike the limit of the sinc function for which all the standard proofs seem to require geometry or calculus.)

Using $cos \ h=4 cos^{3} (\frac{h}{3})-3 cos (\frac{h}{3})$, we get $l= \lim\limits_{h\to 0} \frac{4cos^{3}(\frac{h}{3})-3 cos(\frac{h}{3})-4+3}{h}=\lim\limits_{h\to 0}\frac{4}{3} \frac {(cos (\frac{h}{3})-1)}{\frac{h}{3}}(cos^{2}(\frac{h}{3})+cos(\frac{h}{3})+1) -\lim\limits_{h\to 0}\frac{cos(\frac{h}{3})-1}{\frac{h}{3}}$, and so $l=\frac{4}{3}.3.l -l$ and thus $l=0$.

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  • $\begingroup$ Or just using $cos \ h=2 cos^{2} (h/2)-1$ gives $l=2l$ for the limit $l$. $\endgroup$ – Aritro Pathak Mar 10 '17 at 5:36
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You can generalize the problem a bit. Let $f(h)$ be a function satisfying:

  • $f(0)$=0,
  • $f(h)=f(-h)$,
  • $f(h)$ is Taylor-expandable at $h=0$. (Or $f(h)$ is second-order differentiable at $h=0$.)

Proof: $\lim_{h \rightarrow 0} \frac{f(h)}{h} = 0$.

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    $\begingroup$ You think someone at this level knows what "Taylor-expandable" means? $\endgroup$ – zhw. Jul 6 '16 at 23:16
  • $\begingroup$ He/She just need advance a little bit to understand this generalization. Otherwise, students just get entangled and distracted by the details of a specific function, but cannot see through at the essence. Intuition and concept are more important than technical details. I should replace "Taylor-expandable" with "Second order differentiable". Then the students shall be able to prove with the definition of derivatives and limits. $\endgroup$ – Junning Li Jul 7 '16 at 0:35
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Rewrite:

$$\lim_{h\to 0} \frac{\cos (h)-1}{h}\\\stackrel{1}{=}\lim_{h\to 0}( -\frac{2 \sin^2(h/2)}{h})\\\stackrel{2}{=}-\lim_{\theta \to 0}\frac{\sin \theta}{\theta} \sin \theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Let $\theta=h/2$)} \\ \stackrel{3}{=} -(1)(0)\\\stackrel{4}{=}0$$


  1. Use half-angle identity

  1. Let $\theta = h/2$. Then $h = 2\theta$.

$$\lim_{h\to 0}( -\frac{2 \sin^2(h/2)}{h})$$

$$ = \lim_{h\to 0}( -\frac{2 \sin^2(2\theta/2)}{2\theta})$$

$$ = \lim_{h\to 0}( -\frac{ \sin^2(\theta)}{\theta})$$

$$ = \lim_{\color{red}{\theta}\to 0}( -\frac{ \sin^2(\theta)}{\theta})$$

The last part is because $\theta \to 0$ as $h \to 0$

because $\theta = h/2$ and $h/2 \to 0$ as $h \to 0$


  1. When are we allowed to say that

$$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$$

?

If all the limits involved exist, we are allowed to say that

Remember, not all limit expressions actually make sense exactly. For example

$$\lim_{x \to 0} \frac{|x|}{x}$$

doesn't exactly make sense because as $x \to 0^{+}$, $\frac{|x|}{x}$ approaches a different value from when $x \to 0^{+}$.

So when we're told to evaluate some limit expression $\lim_{x \to a} f(x)g(x)$, we ought to be told that we're assuming such expression makes sense (the limit exists).

So assuming $$\lim_{\theta\to 0}( -\frac{ \sin^2(\theta)}{\theta})$$ exists, evaluate it.

We know that the following limits exist

$$\lim_{\theta\to 0}-\frac{ \sin(\theta)}{\theta}$$

$$\lim_{\theta\to 0}\sin(\theta)$$

Therefore assuming $$\lim_{\theta\to 0}( -\frac{ \sin^2(\theta)}{\theta})$$ exists, we are allowed to say that

$$\lim_{\theta\to 0}( -\frac{ \sin^2(\theta)}{\theta})$$

$$= \lim_{\theta\to 0}-\frac{ \sin(\theta)}{\theta} \lim_{\theta\to 0}\sin(\theta)$$


  1. $$0 \times -1=0$$
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I'm going to explain your proof first of all your proof used half angle formula:
$\cos(x)=1-\sin^2(\frac{x}{2})$
This formula comes from other formulas I hope you know them:
f1: $\sin^2(x)+\cos^2(x)=1$
and
f2: $\cos(2x)=\cos^2(x)-\sin^2(x)$
Look at f1 we can change it to both:
f3: $\cos^2(x) = 1 - \sin^2(x)$
f4: $\sin^2(x) = 1 - \cos^2(x)$

If we substitute f3 inside f2 we will have:
f5: $\cos(2x) = 1 - \sin^2(x) - \sin^2(x) = 1 - 2\sin^2(x)$
So if: $2x = h$ and $x = \frac{h}{2} $ Then:
f6: $\cos(h) = 1-2\sin^2(h/2) $

Let's get back to: $\lim\limits_{h\to0}\frac{\cos(h)-1}{h}$ If we substitute f6 in it we will have $\lim\limits_{h\to0}\frac{ 1-2\sin^2(h/2) -1}{h}$ Now you we just clean our result by turning $-2\sin(h/2)$ to $-\sin(h/2)$:
$\lim\limits_{h\to0}\frac{ \frac {-2\sin^2(h/2)}{2} }{\frac{h}{2}}= \lim\limits_{h\to0}\frac{-\sin^2(h/2)}{h/2}$
And making our result more clear:
$ \lim\limits_{h\to0}\frac{-1×\sin(h/2)×\sin(h/2)}{h/2} = \lim\limits_{h\to0}-1×\sin(h/2)×\frac{\sin(h/2)}{h/2} $ The problem is now: $ \frac{\sin(h/2)}{h/2}$ Check out why $ \lim\limits_{x\to0} \frac{\sin(x)}{x}=1$ ?

So now we have: $-1×1×0=0$

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For convenience, let us double the argument:

$$\lim_{h\to0}\frac{\cos h-1}h=\lim_{2h\to0}\frac{\cos2h-1}{2h}=\lim_{h\to0}\frac{\cos2h-1}{2h}.$$

Then it is well-known that $\cos2h=1-2\sin^2h$ and we have

$$-\lim_{h\to0}\frac{2\sin^2h}{2h}=-\lim_{h\to0}\frac{\sin h\sin h}{h}=-\lim_{h\to0}\frac{\sin h}{h}\lim_{h\to0}\sin h=-1\cdot0.$$

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