Cfrgtkky

You should first prove that for $x > 0$ small that $sin x < x < tan x$. Then, dividing by $x$ you get
$$
{ sin x over x} < 1
$$
and rearranging $1 < {tan x over x} = {sin x over x cos x }$
$$
cos x < {sin x over x}.
$$
Taking $x rightarrow 0^+$ you apply the squeeze theorem. For $x < 0$ and small use that $sin(-x) = -sin x$ so that $${sin(-x) over -x} = {sin x over x}.$$
As far as why the first inequality I said is true, you can do this completely from triangles but I don't know how to draw the pictures here.

Usually calculus textbooks do this using geometric arguments followed by squeezing.

Here's an Euler-esque way of looking at it---not a "proof" as that term is usually understood today, but still worth knowing about.

Let $theta$ be the length of an arc along the circle of unit radius centered at $(0,0)$, from the point $(1,0)$ in a counterclockwise direction to some point $(costheta,sintheta)$ on the circle. Then of course $sintheta$ is the height of the latter point above the $x$-axis. Now imagine what happens if $theta$ is an infinitely small positive number. Then the arc is just an infinitely short vertical line, and the height of the endpoint above the $x$-axis is just the length of the arc. I.e. when $theta$ is an infinitely small number, then $sintheta$ is the same as $theta$. It follows that when $theta$ is an infinitely small nonzero number, then $dfrac{sintheta}{theta}=1$.

That is how Euler viewed the matter. See his book on differential calculus.

Look at this link:

http://fatosmatematicos.blogspot.com/2010/08/provas-sem-palavras-parte-20.html

Here is the picture I copied from that blog:

Here you may see an elementary approach that starts from a very interesting result, see this problem. All you need is a bit of imagination. When we take $lim_{nrightarrowinfty} frac{nsin(frac{pi}{n})}{1+sin(frac{pi}{n})}$ we may notice that we have infinitely many circles surrounding the unit circle with infinitely small diameters that lastly perfectly approximate the length of the unit circle when having it there infinity times . Therefore when multiplying n by the radius under the limit to infinity we get π. Let's denote $frac{pi}{n}$ by x.

$$lim_{xrightarrow0}frac{pisin(x)}{x(1+sin(x))}={pi}Rightarrowlim_{xrightarrow0}frac{sin(x)}{x(1+sin(x))}=1Rightarrowlim_{xrightarrow0}frac{sin(x)}{x}=1$$

The proof is complete.

I claim that for $0<x<pi/2$ that the following holds
$$sin x lt x lt tan x$$


In the diagram, we let $OC=OA=1$. In other words, $Arc:CA=x$ is an arc of a unit circle. The shortest distance from point $C$ to line $AO$ is line $CE=sin x$ (because $CEperp OA$). Another path from point $C$ to line $OA$ is arc $CA$ (which is longer than CE because it is not the shortest path). So we have at the very least
$$sin x lt x$$
Now we need to show that line $BA=tan x gt x$.

Lines $AD$ and $CD$ are both tangent to arc $CA$. $CD+DA$ is longer than arc $CA$ because the set of points bound by sector $OCA$ is a subset of the set of points bound by quadrilateral $OCDA$, both of which are convex sets. This means that the perimeter of quadrilateral $OCDA$ must be longer than the perimeter of the sector $OCA$ (as per Archimedes, On the Sphere and Cylinder Book I). But both the sector and the quadrilateral both have sides $OC$ and $OA$, so we have
$$CA=x<DC+DA$$
But $BD>CD$ because it is the hypotenuse in $triangle BCD$ we have
$$tan x = BA = BD+DAgt CD+DA gt CA=x gt sin x$$

So we have
$$sin x lt x lt tan x$$
$$frac{sin x}{x} lt 1 lt frac{tan x}{x}=frac{sin x}{x}cdotsec x$$
From this we can extract
$$frac{sin x}{x} lt 1$$
and
$$1 lt frac{sin x}{x}cdotsec x$$
$$cos x lt frac{sin x}{x}$$
Putting these inequalities back together we have
$$cos x lt frac{sin x}{x} lt 1$$

Because $displaystylelim_{xto 0}cos x = 1$, by the squeeze theorem we have
$$lim_{xto 0}frac{sin x}{x}=1$$

I am not sure if it counts as proof, but I have seen this done by a High Schooler.

In the given picture above,
$displaystyle 2n text{ EJ} = 2nR sinleft( frac{pi}{n } right ) = text{ perimeter of polygon }$.

$displaystyle lim_{nto infty }2nR sinleft( frac{pi}{n } right ) = lim_{nto infty } (text{ perimeter of polygon }) = 2 pi R implies lim_{nto infty}frac{sinleft( frac{pi}{n } right )}{left( frac{pi}{n } right )} = 1$ and let $frac{pi}{n} = x$.

Don't you feel strange about why most of the proofs are done with a figure? I've had this problem in the beginning, and realized after that this is due to the definition we use for the function $sin x$. Because the usual definition of $sin x$ we all study first in high schools depends on “classical geometry” and usually with a figure, you should depict out the figure and to make it clear.

However, if you use other definitions of $sin x$ that are equivalent to the former, you'll find it more simple. For example,

$$sin x = frac{x^1}{1!} - frac{x^3}{3!}+ frac{x^5}{5!} - cdots + cdots - cdots$$

and hence

$$frac{sin x}x = frac{x^0}{1!} - frac{x^2}{3!}+ frac{x^4}{5!} - cdots$$

which obviously tends to $1$ as $x$ approaches 0.

Here's one more:
$$
lim_{x to 0} frac{sin x}{x}=lim_{x to 0} lim_{v to 0}frac{sin (x+v)-sin v}{x}\
=lim_{v to 0} lim_{x to 0}frac{sin (x+v)-sin v}{x}=lim_{v to 0}sin'v=lim_{v to 0} cos v=1
$$

It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.

The strategy is to find $frac{darcsin y}{dy}$ first. This can easily be done using the picture below.

From the above picture, $arcsin y$ is twice the area of the orange bit. The area of the red bit is ${1 over 2}ysqrt{1-y^2}$. The area of the red bit plus the orange bit is $int_{0}^y sqrt{1-Y^2} dY$. So $$arcsin y = 2int_{0}^y sqrt{1-Y^2} dY - ysqrt{1-y^2}.$$ Differentiating with respect to $y$ gives $frac{darcsin y}{dy} = frac{1}{sqrt{1-y^2}}$. Using the theorem for the derivative of inverse functions yields $sin' theta = sqrt{1 - sin^2 theta} = cos theta$.

(A similar thing can be done with the arc length definition of $arcsin$.)

Let $f:{yinmathbb{R}:yneq 0}tomathbb{R}$ be a function defined by $f(x):=dfrac{sin x}{x}$ for all $xin {yinmathbb{R}:yneq 0}$.

We have $displaystylelim_{x to 0}dfrac{sin x}{x}=1$ if and only if for every $varepsilon>0$, there exists a $delta>0$ such that $|f(x)-1|<varepsilon$ whenever $0<|x-0|<delta$.

Let $varepsilon>0$ be an arbitrary real number.

Note that $sin x=displaystyle sum_{n=0}^{infty}(-1)^ndfrac{x^{2n+1}}{(2n+1)!}$.

If $x neq 0$, we have $dfrac{sin x}{x}=$$displaystyle sum_{n=0}^{infty}(-1)^ndfrac{x^{2n}}{(2n+1)!}=1+$$displaystyle sum_{n=1}^{infty}(-1)^ndfrac{x^{2n}}{(2n+1)!}$.

We thus have

$|f(x)-1|=left|dfrac{sin x}{x}-1right|=left|displaystyle sum_{n=1}^{infty}(-1)^ndfrac{x^{2n}}{(2n+1)!}right|leq left|displaystylesum_{n=1}^{infty} dfrac{x^{2n}}{(2n+1)!}right|leq displaystylesum_{n=1}^{infty} left|dfrac{x^{2n}}{(2n+1)!}right|$

Therefore we have

$|f(x)-1|leq displaystylesum_{n=1}^{infty} left|dfrac{x^{2n}}{(2n+1)!}right|leq displaystyle sum_{n=1}^{infty} |x^{2n}|=sum_{n=1}^{infty}|x^2|^n$

If $0<|x|<1$, then $0<|x^2|<1$, and the infinite series $displaystylesum_{n=1}^{infty}|x^2|^n$ converges to $dfrac{x^2}{1-x^2}$.

Choose $delta:=sqrt{dfrac{varepsilon}{1+varepsilon}}$.
Then $0<|x-0|<delta$ implies that $0<|x|<$$sqrt{dfrac{varepsilon}{1+varepsilon}}<1$, and hence $x^2<varepsilon-varepsilon x^2$. But $x^2<varepsilon-varepsilon x^2$ implies that $dfrac{x^2}{1-x^2}<varepsilon$.

We therefore have $sum_{n=1}^{infty}|x^2|^n<varepsilon$ whenever $0<|x-0|<delta$. But since $|f(x)-1|leqdisplaystylesum_{n=1}^{infty}|x^2|^n$, we have $|f(x)-1|<varepsilon$ whenever $0<|x-0|<delta$.

Since $varepsilon$ was arbitrary, we have $displaystylelim_{x to 0}dfrac{sin x}{x}=1$.

Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest without appeal to geometry or differential calculus.

Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.

We define the sine function, $sin(x)$, as the inverse function of the function $f(x)$ given by

$$bbox[5px,border:2px solid #C0A000]{f(x)=int_0^x frac{1}{sqrt{1-t^2}},dt }tag 1$$

for $|x|< 1$.

NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $sin(x)$.

It is straightforward to show that since $frac{1}{sqrt{1-t^2}}$ is positive and continuous for $tin (-1,1)$, $f(x)$ is continuous and strictly increasing for $xin (-1,1)$ with $displaystylelim_{xto 0}f(x)=f(0)=0$.

Therefore, since $f$ is continuous and strictly increasing, its inverse function, $sin(x)$, exists and is also continuous and strictly increasing with $displaystyle lim_{xto 0}sin(x)=sin(0)=0$.

From $(1)$, we have the bounds (SEE HERE)

$$bbox[5px,border:2px solid #C0A000]{1 le frac{f(x)}xle frac{1}{sqrt{1-x^2}}} tag 2$$

for $xin (-1,1)$, whence applying the squeeze theorem to $(2)$ yields

$$lim_{xto 0}frac{f(x)}{x}=1 tag 3$$

Finally, let $y=f(x)$ so that $x=sin(y)$. As $xto 0$, $yto 0$ and we can write $(3)$ as

$$lim_{yto 0}frac{y}{sin(y)}=1$$

from which we have

$$bbox[5px,border:2px solid #C0A000]{lim_{yto 0}frac{sin(y)}{y}=1}$$

as was to be shown!

NOTE:

We can deduce the following set of useful inequalities from $(2)$. We let $x=sin(theta)$ and restrict $x$ so that $xin [0,1)$. In addition, we define new functions, $cos(theta)=sqrt{1-sin^2(theta)}$ and $tan(theta)=sin(theta)/cos(theta)$.

Then, we have from $(2)$

$$bbox[5px,border:2px solid #C0A000]{ycos(y)le sin(y)le yle tan(y)} $$

which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.

Because $sin x$ has zeroes at $x=n pi$ for arbitrary integer $n$ including $x=0$, you can use Vieta's Theorem $sin x = A(cdots(x+2 pi)(x+pi)x(x-pi)(x-2 pi)cdots)$ with a constant $A$. Because $sin(frac{pi}{2})=1$ this constant can be determined by the equation: $$1=A(cdots(frac{pi}{2}+2 pi)(frac{pi}{2}+pi)frac{pi}{2}(frac{pi}{2}-pi)(frac{pi}{2}-2 pi)cdots).$$

Now, in the Expression $f(x):= frac{sin(x)}{x}$ the $x$ cancels such that $$f(x)=A(cdots(x+2 pi)(x+pi)(x-pi)(x-2 pi)cdots),$$ hence: $$lim_{x rightarrow 0} f(x)=A(cdots(2 pi) cdot picdot(- pi)cdot(-2 pi)cdots) = A prod_{k=1}^infty (-k^2 pi^2).$$

$frac{1}{A} = frac{pi}{2} prod_{k=1}^infty ((frac{pi}{2})^2-k^2 pi^2)$. The proof is completed when the Wallis product is used.

Usual proofs can be circular, but there is a simple way for proving such inequality.

Let $theta$ be an acute angle and let $O,A,B,C,D,C'$ as in the following diagram:

We may show that:

$$ CD stackrel{(1)}{ geq };stackrel{largefrown}{CB}; stackrel{(2)}{geq } CB,stackrel{(3)}{geq} AB $$

$(1)$: The quadrilateral $OCDC'$ and the circle sector delimited by $O,C,C'$ are two convex sets. Since the circle sector is a subset of the quadrilateral, the perimeter of the circle sector is less than the perimeter of the quadrilateral.

$(2)$: the $CB$ segment is the shortest path between $B$ and $C$.

$(3)$ $CAB$ is a right triangle, hence $CBgeq AB$ by the Pythagorean theorem.

In terms of $theta$ we get:
$$ tantheta geq theta geq 2sinfrac{theta}{2} geq sintheta $$
for any $thetainleft[0,frac{pi}{2}right)$. Since the involved functions are odd functions the reverse inequality holds over $left(-frac{pi}{2},0right]$, and $lim_{thetato 0}frac{sintheta}{theta}=1$ follows by squeezing.

A slightly different approach might be the following one: let us assume $thetainleft(0,frac{pi}{2}right)$.
By $(2)$ and $(3)$ we have
$$ theta geq 2sinfrac{theta}{2}geq sintheta $$
hence the sequence ${a_n}_{ngeq 0}$ defined by $a_n = 2^n sinfrac{theta}{2^n}$ is increasing and bounded by $theta$. Any increasing and bounded sequence is convergent, and we actually have $lim_{nto +infty}a_n=theta$ since $stackrel{largefrown}{BC}$ is a rectifiable curve and for every $ngeq 1$ the $a_n$ term is the length of a polygonal approximation of $stackrel{largefrown}{BC}$ through $2^{n-1}$ equal segments. In particular

$$ forall thetainleft(0,frac{pi}{2}right), qquad lim_{nto +infty}frac{sinleft(frac{theta}{2^n}right)}{frac{theta}{2^n}} = 1 $$
and this grants that if the limit $lim_{xto 0}frac{sin x}{x}$ exists, it is $1$. By $sin xleq x$ we get $limsup_{xto 0}frac{sin x}{x}leq 1$, hence it is enough to show that $liminf_{xto 0}frac{sin x}{x}geq 1$. We already know that for any $x$ close enough to the origin the sequence $frac{sin x}{x},frac{sin(x/2)}{x/2},frac{sin(x/4)}{x/4},ldots$ is convergent to $1$, hence we are done.

Long story short: $lim_{xto 0}frac{sin x}{x}=1$ follows from the fact that a circle is a rectifiable curve, and a circle is a rectifiable curve because it is the boundary of a convex, bounded subset of $mathbb{R}^2$. The convexity of the disk follows from the triangle inequality: a disk is a closed ball for the euclidean distance.

$(1)$ relies on this powerful Lemma:

Lemma. If $A,B$ are convex bounded sets in $mathbb{R}^2$ and $Asubsetneq B$, the perimeter of $A$ is less than the perimeter of $B$.

Proof: by boundedness and convexity, $partial A$ and $partial B$ are rectifiable, with lengths $L(A)=mu(partial A),,L(B)=mu(partial B)$. Always by convexity, there is some chord in $B$ that does not meet the interior of $A$ (a tangent to $partial A$ at a smooth point does the job, for instance). Assume that such chord has endpoints $B_1, B_2 in partial B$ and perform a cut along $B_1 B_2$: both the area and the perimeter of $B$ decrease, but $B$ remains a bounded convex set enclosing $A$. Since $A$ can be approximated through a sequence of consecutive cuts, $L(A)<L(B)$ follows.

Let $sin(x)$ is defined as solution of $frac{d^2}{dx^2}textrm{f}(x)=-textrm{f}(x)$ with $mathrm f(0)=0,,frac{d}{dx}mathrm f(0)=C$ initial conditions, so exact solution is $mathrm f(x)=Ccdotsin(x)$.
Define second derivative as
$$
begin{align*}
frac{d^2}{dx^2}textrm{f}(x)=lim_{Delta xto 0}{frac{frac{mathrm f(x)-mathrm f(x-Delta x)}{Delta x}-frac{mathrm f(x-Delta x)-mathrm f(x-2cdotDelta x)}{Delta x}}{Delta x}}&=\=lim_{Delta xto 0}{frac{mathrm f(x)-2cdot mathrm f(x-Delta x)+mathrm f(x-2cdotDelta x)}{Delta x^2}}
end{align*}
$$
we can easy check this limit for any (?) functions. Similarly, we can define the first derivative for right, middle and left points:
$$
frac{d}{dx}textrm{f}(x)left{
begin{aligned}
&=lim_{Delta xto 0}{frac{mathrm f(x)-mathrm f(x-Delta x)}{Delta x}}
\
&=lim_{Delta xto 0}{frac{mathrm f(x-Delta x)-mathrm f(x-2cdotDelta x)}{Delta x}}\
&=lim_{Delta xto 0}{frac{mathrm f(x)-mathrm f(x-2cdotDelta x)}{2cdotDelta x}}
end{aligned}
right.
$$
Let's use the finite elements method assuming $Td=Delta x,,y_n=mathrm f(x),,y_{n-1}=mathrm f(x-Delta x),,y_{n-2}=mathrm f(x-2cdot Delta x)$
Override differential equation as
$$
frac{y_n-2cdot y_{n-1}+y_{n-2}}{Td^2}=-y_n
$$
Now solve this implicit equation for $y_n$ to obtain explicit recurrence relation:
$$
y_n = frac{2cdot y_{n-1}-y_{n-2}}{1+Td^2}
$$
Using arbitrarily small but non-zero quantity Td we can plot exponentially decaying sampled sine function (because the poles are inside the unit circle of the transfer function corresponding to the given recurrence relation).
Similarly we write three systems for the initial conditions:

$$
left{
begin{aligned}
&y_n = frac{2cdot y_{n-1}-y_{n-2}}{1+Td^2}
\
&C=frac{y_n-y_{n-1}}{Td}
end{aligned}right.
$$
$$
left{
begin{aligned}
&y_n = frac{2cdot y_{n-1}-y_{n-2}}{1+Td^2}
\
&C=frac{y_{n-1}-y_{n-2}}{Td}
end{aligned}right.
$$
$$
left{
begin{aligned}
&y_n = frac{2cdot y_{n-1}-y_{n-2}}{1+Td^2}
\
&C=frac{y_n-y_{n-2}}{2cdot Td}
end{aligned}right.
$$
Solve this sequence of equations for $y_{n-1}$ and $y_{n-2}$:
$$
left{
begin{aligned}
&y_{n-1} = -Ccdot Td + y_{n}\
&y_{n-2}=-2cdot Ccdot Td + y_{n}cdotleft(1-Td^2right)
end{aligned}right.
$$
$$
left{
begin{aligned}
&y_{n-1} = -Ccdot Td + y_{n}cdotleft(1+Td^2right)\
&y_{n-2}=-2cdot Ccdot Td + y_{n}cdotleft(1+Td^2right)
end{aligned}right.
$$
$$
left{
begin{aligned}
&y_{n-1} = -Ccdot Td + y_{n}cdotleft(1+frac{Td^2}{2}right)\
&y_{n-2}=-2cdot Ccdot Td + y_{n}
end{aligned}right.
$$
At zero point $y_n=mathrm f(0)=0$ and we can see linear dependence:
$$
begin{aligned}
&y_{n-1} = -Ccdot Td\
&y_{n-2}=-2cdot Ccdot Td
end{aligned}
$$
for all three solutions. Replace back:
$$
begin{array}{l}
mathrm f(0)&=0\
mathrm f(0-Delta x) &= -Ccdot Delta x\
mathrm f(0-2cdot Delta x) &= -2cdot Ccdot Delta x
end{array}
$$
So all three $frac{d}{dx}mathrm f(0)$ limits is equal to $C$ at $x=0$ and in accordance with $mathrm f(x)=Ccdotsin(x)$ by definition we can write
$$
lim_{Delta xto 0}{frac{mathrm f(0)-mathrm f(0-Delta x)}{Delta x}}=lim_{Delta xto 0}{frac{0-(-C cdot Delta x)}{Delta x}}=C
$$
Thus
$$
lim_{Delta xto 0}{frac{sin(0)-Ccdotsin(0-Delta x)}{Delta x}}=lim_{Delta xto 0}{frac{Ccdotsin(Delta x)}{Delta x}}=Ccdotlim_{Delta xto 0}{frac{sin(Delta x)}{Delta x}}=C
$$
and $lim_{Delta xto 0}{frac{sin(Delta x)}{Delta x}}=1$

Simple one is using sandwich theorem Which demonstrated earlier.In this method you have to show that $frac{sin x}{x} $ lies between other two functions. As $x longrightarrow 0$ both of them will tends to ONE.

Then as in the case of sandwich (if both the bread part go to one stomach the middle portion will also go to the same stomach) $frac{sin x}{x}$ will go to ONE.

You can use geogebra to see the visualization of this phenomena using geogebra.First you input $sin x$ and $x$ and observe that near to $0$ values of $sin x$ and $x$ are same.

Secondly input $frac{sin x}{x}$ then observe function is approaching to $1$ as $x$ tends to $0$

Originally posted on the proofs without words post, here is a simple image that explains the derivative of $sin(x)$, which as we all know, is directly related to the limit at hand.

If one is not so convinced, take a look at the above picture and notice that if $upm h$ is in the first quadrant, then

$$frac{sin(x+h)-sin(x)}h<cos(x)<frac{sin(x-h)-sin(x)}{-h}$$

Notice that

$$
begin{align}frac{d}{dx}sin(x)&=lim_{hto0}frac{sin(x+h)-sin(x)}h\text{picture above}&=lim_{hto0}frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}h\cos(x)&=lim_{hto0}sin(x)frac{cos(h)-1}h+cos(x)frac{sin(h)}h\cos(0)&=lim_{hto0}frac{sin(h)}hend{align}
$$

Here is another approach.

(1)
(2)

In the large triangle, $$tan(theta)=frac{opp}{adg}=frac{z}{1}=z$$ So the triangle has height $$z=tan(theta)$$ and base $1$
so it's area is $$Area(big triangle)=frac{1}{2}z=frac{1}{2}tan(theta)$$

Next the sector area as a fraction of the entire circle, the sector is (see the right hand side of picture (1))$$frac{theta}{2pi}$$ of the entire circle so it's area is

$$Area(sector)=frac{theta}{2pi}*(pi)(1)^2=frac{theta}{2}$$
The triangle within the sector has height $y$. But $y=frac{y}{1}=frac{opp}{hyp}=sin(theta)$ so the small triangle has height $y=sin(theta)$ and base $1$ so its area is $$Area(small triangle)=frac{1}{2}y=frac{1}{2}sin(theta)$$
Now we can use the sandwich theorem as $$Area(big triangle)geq Area(sector)geq Area(small triangle)$$

using the equations we worked out above this becomes

$$frac{tan(theta)}{2}geqfrac{theta}{2}geqfrac{sin(theta)}{2}$$ now multipliying through by two and using the fact that $$tan(theta)=frac{sin(theta)}{cos(theta)}$$ we get that
$$frac{sin(theta)}{cos(theta)}geqthetageqsin(theta)$$
taking reciprocals changes the inequalities so we have that
$$frac{cos(theta)}{sin(theta)}leqfrac{1}{theta}leqfrac{1}{sin(theta)}$$
now finally multiplying through by $sin(theta)$ we get $$cos(theta)leqfrac{sin(theta)}{theta}leq1$$
Now $$lim limits_{theta to 0}cos(theta)=1$$ and$$lim limits_{theta to 0}1=1$$

so by the sandwhich theorem $$lim limits_{theta to 0}frac{sin(theta)}{theta}=1$$ also. QED

We may use a linear approximation for this limit. The following expression comes straight out of the equation of a line that is $y=mx+b$. Using the fact that $m=dfrac{d}{dx}[f(x)]$ $$f(x) approx f(a) + f'(a)(x-a)$$In this case $f(x)=sin x$ and $a=0$ $implies sin x approx f(0)+cos 0 (x-0)=x longleftrightarrow sin x approx x$. The following graph better illustrates this tangent line approximation.

Since, $sin x approx x implies displaystyle lim_{x to 0} dfrac{sin x}{x}=1$. Quod Erat Demonstrandum

Here is a proof to those familiar with power series.

The definition of $sin(x)$ is

$$sin(x) = sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}h^{2k+1}$$

Therefore we get

$$begin{align} lim_{x to 0} frac{sin(x)}{x} &= lim_{x to 0} frac{sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}x^{2k+1}}{x} \&= lim_{x to 0} sum_{k=0}^infty frac{(-1)^k}{(2k+1)!} x^{2k} \&= 1 + lim_{x to 0} sum_{k=1}^infty frac{(-1)^k}{(2k+1)!} x^{2k} \&= 1 end{align}$$

where we have used the fact that the power series $sum_{k=1}^infty frac{(-1)^k}{(2k+1)!} x^{2k}$ has radius of convergence $R=infty$ and therefore is continuous on $mathbb R$. This allows us to take the limit inside and we get

$$lim_{x to 0} sum_{k=1}^infty frac{(-1)^k}{(2k+1)!} x^{2k} = sum_{k=1}^infty frac{(-1)^k}{(2k+1)!} 0^{2k} = 0$$

We can also use Euler's formula to prove the limit:

$$e^{ix} = cos x + isin x$$

$$lim_{xto 0}dfrac{sin x}{x} = implies lim_{xto 0} dfrac{e^{ix}- e^{-ix}}{2i x}$$

$$= lim_{xto 0} dfrac{e^{2ix}-1}{2ix}timesdfrac 1{ e^{ix} }= 1 times 1 = 1$$

since:

$lim_{f(x)to 0}dfrac{e^{f(x)}-1}{f(x)} = 1$