Understanding this proof that $\lim\limits_{h\to 0}\frac{\cos(h)-1}{h}=0$

Question

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!

Answer 1

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

Answer 2

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 $$ \lim_{h\to 0} \frac{\cos(h)-1}{h} = -\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}\cdot \sin(\theta) = -\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} \cdot \lim_{\theta \to 0} \sin(\theta) = - 1 \cdot 0 = 0. $$

Answer 3

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 $l=0$.

Answer 4

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

Answer 5

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

---

2. 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 $h \to 0$

---

3. 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

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

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

---

4. $$0 \times -1=0$$

Answer 6

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: $-1×1×0=0$

Answer 7

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

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