## Limits of functions Lecture 8

### Topic: Limit of a function \underset{x\rightarrow0}{\lim}\frac{sinx}{x}=1 – proof

#### Summary

In the article I will show proof that:

\underset{x\rightarrow0}{\lim}\frac{sinx}{x}=1I didn’t come up with the proof myself, I just took it from GMFichtenholz’s book “Differential and Integral Calculus” (with a small modification at the very end) – of course extending it to include all transitions, for a better understanding of what comes from what. To demonstrate the above limit, I will use Cauchy’s definition of the limit of a function.

#### Demonstrating a certain inequality useful for proof

**Lemma**

*For 0<x<\frac{1}{2} :*

*sinx<x i x<tgx .*

**Proof of the lemma**

So we our task is to prove the inequality:

sinx< x< tgx, which occurs for *0< x< \frac{1}{2}*

This inequality is worth remembering, because it is useful not only for this proof, but for various other things in mathematical analysis (for example, for estimating numerical series in a comparative convergence criterion). Besides, it’s interesting in itself, right? 🙂

Let’s go ahead and prove this inequality.

First, let’s draw something like this:

We have a circle with center O, radius R, chord AB, angle AOB equal to \alpha , and tangent BC to the circle at point B.

Let’s note three things:

- Area of triangle OAB
- Area of a sector of a circle with angle \alpha AOB (do not confuse it with the triangle from point 1!)
- Area of triangle COB

It is obvious that always:

Area of triangle OAB < AOB circle’s slice field < Area of triangle COB

Now let’s compute these fields one by one:

1. Area of triangle OAB

From high school we remember an alternative to the clichéd ” P=\frac{1}{2}ah ” formula:

*Area of a triangle = \frac{1}{2} side length times side length times sine of the angle between them.*

In our particular triangle, both sides have length R (radius of the circle), so the area of the triangle is:

P_{\Delta} OAB=\frac{1}{2}{R}^{2}sin{\alpha}2. OAB slice field

The area of the sector can be determined, for example, from proportions (or a ready-made formula – if you know it). The area of the entire circle corresponds to the angle 2\pi (in radians), and the area of the segment (which we are about to calculate) corresponds to the angle \alpha radians. The proportions will look like this:

{\pi}{R}^{2} — 2\pi

P(AOB) — \alpha

therefore:

\frac{{\pi}{R}^{2}}{P(AOB)}=\frac{2\pi}{\alpha}By cross-multiplying, as in proportions (we are looking for P(AOB)), we will get our sector area:

P(AOB)=\frac{{\pi} R^2{\alpha} }{2\pi} =\frac{R^2{\alpha} }{2}=\frac{1}{2}R^2{\alpha}

3. Area of triangle COB

This triangle is a right triangle (because line CB is a tangent line). Therefore, we can assume that the segment OB (equal to R) is its base, and the segment CB – the height, and use the standard P = \frac{1}{2}ah formula.

P_{\Delta} COB=\frac{1}{2}|OB||CB|=\frac{1}{2}R|CB|

However, let’s include angle \alpha in this formula, instead of |CB| . We know that in this triangle:

tg{\alpha} =\frac{|CB|}{|OB|}=\frac{|CB|}{R}

Multiplying both sides of the equality by R we get:

|CB|=Rtg{\alpha}

…which we can substitute into the formula for the area of triangle COB, obtaining:

P_{\Delta} COB=\frac{1}{2}|OB||CB|=\frac{1}{2}R\cdot{Rtg{\alpha}} =\frac{1}{2}R^2{tg{\alpha}}

In this way, we have all 3 areas from the picture above computed.

So now let’s go back to inequality:

Area of triangle OAB < AOB slice field < Area of triangle COB

Substituting the calculated fields we get:

\frac{1}{2}R^2 sin{\alpha}< \frac{1}{2}R^2{\alpha}< \frac{1}{2}R^2tg{\alpha}We divide both sides by \frac{1}{2}R^2 (and we can do so, because the radius of the circle is greater than zero) and we get:

sin{\alpha}<{\alpha}< tg{\alpha}And this is exactly the inequality we wanted to prove! And we have just proved this way 🙂

#### Conclusions from the proven inequality

So let’s start with the already proven inequality and transform it a bit:

sinx < x < tgxFirst, let’s break it down into two ineqalities:

1. sinx < x

We divide it on both sides by x (we can, because x is assumed to be 0 < x < {pi}/2 – i.e. non-zero and positive, so the inequality sign will not change):

\frac{sinx}{x} < 12. x < tgx

As we know tgx=\frac{sinx}{cosx} :

x < \frac{sinx}{cosx}After multiplying by cosx (we can, due to the assumption about x’s mentioned earlier):

xcosx < sinxAnd divided by x (we can, due to the assumption about x):

cosx<\frac{sinx}{x}So in this way, we have two inequalities transformed:

\frac{sinx}{x} < 1 cosx < \frac{sinx}{x}Which we can combine and write:

cosx < \frac{sinx}{x} < 1After subtracting 1 by both sides we get (if you are confused, you can break it down into two inequalities):

cosx-1 < \frac{sinx}{x}-1 < 0We multiply both sides by -1 (this will change the inequality signs):

We write more neatly:

0 < 1-\frac{sinx}{x} < 1-cosxAt first, let’s deal with 1-cosx :

At the beginning, unfortunately, we must repeat a certain trigonometric formula, namely the formula for the cosine of a double angle:

cos2x=cos^2{x} -sin^2{x}So (according to this formula):

cosx=cos^2(\frac{1}{2} x)-sin^2(\frac{1}{2} x)That is:

1-cosx=1-(cos^2(\frac{1}{2} x)-sin^2(\frac{1}{2} x))By using the “Pythagorean identity” I get from this:

1-cosx=cos^2(\frac{1}{2} x)+sin^2(\frac{1}{2} x)-cos^2(\frac{1}{2} x)+sin^2(\frac{1}{2} x)That is:

1-cosx=2sin^2(\frac{1}{2} x)So back to our inequality:

0 < 1-\frac{sinx}{x} < 1-cosxWe can write it as:

0 < 1-\frac{sinx}{x} < 2sin^2(\frac{1}{2} x)Now let’s deal with 2sin^2(\frac{1}{2} x) .

It is true that:

2sin^2(\frac{1}{2} x) < 2sin(\frac{1}{2} x)because for x it is assumed to be in the range 0 < x < \frac{\pi}{2}, in which sine values are fractions in the interval (0,1), and such squared fractions are **smaller** than before squared (for example \frac{1}{16} < \frac{1}{4}).

It is also true that:

2sin(\frac{1}{2} x) < xbecause after dividing the inequality by 2 we get:

2sin(\frac{1}{2} x) < \frac{x}{2}and we’ve already showed in the lemma that the sine of something is smaller than that something ( sinx < x < tgx ) at the beginning.

So going back to our main inequality:

0 < 1-\frac{sinx}{x} < 2sin^2 (\frac{1}{2} x)< 2sin(\frac{1}{2} x) < xAnd making it shorter:

0 < 1-\frac{sinx}{x} < xThe above inequality results in the following inequality:

|1-\frac{sinx}{x} | < | x |– the absolute value is the distance from zero, so it will not change when x changes to negative, as long as x remains between 0 and \frac{\pi}{2} .

Due to the fact that in absolute values | a-b | = | b-a | :

| \frac{sinx}{x} - 1 | < | x |Let us repeat, we have shown that the below inequality holds (for 0 < x < \frac{\pi}{2} ):

| \frac{sinx}{x} - 1 | < | x |#### Cauchy definition of the limit of the function

Recall: if we want to prove from the Cauchy definition of the limit of the function that \underset{x\rightarrow0}{\lim}\frac{sinx}{x}=1 we have to show that we will always find one \delta that for every \epsilon the follow will occur:

0 < |x-0| < \delta\Rightarrow | \frac{sinx}{x} - 1| < \epsilonThat is:

0 < |x| < \delta\Rightarrow | \frac{sinx}{x} - 1| < \epsilonLet’s start by acknowledge simple fact: if we choose just any \epsilon – indeed we can always find some \delta for it as a number greater than zero, less than epsilon and in the range 0 < \delta < \frac{\pi}{2} as well.

Next, from assumption 0 < |x| < \delta our x’s will be smaller than the delta, and since we showed earlier in the proof, that | \frac{sinx}{x}-1 | < | x | we conclude that:

| \frac{sinx}{x}-1 | < | x | < \delta < \epsilonSo in this way we have shown, that for any given \epsilon we can find \delta, that below is true :

| \frac{sinx}{x} -1 | < \epsilonSo we have shown from Cauchy’s definition that:

\underset{x\rightarrow0}{\lim}\frac{sinx}{x}=1

While writing this post, I used…

1. “Differential and integral calculus. Volume I.” GM Fichtenholz. Polish Ed. 1966.

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Click to remind yourself how to apply limits of functions by definition (previous Lecture)< —