*Notes to a video lecture on http://www.unizor.com*

__Amazing Limits__

There are two very important limits that require special attention and non-trivial approach to analyze.

In this lecture we present them with full proof.

*Amazing Limit #1*

**as**

*sin(x)/x → 1*

*x→0**Proof*

We will prove this for angles

*x*in the first quadrant only where

*x*,

*sin(x)*and

*tan(x)*are positive, the general case trivially follows from it. Let's start from the definition of trigonometric functions using the unit circle.

On the picture above

*sin x*segment is an actual definition of a

*sine*. A segment marked

*tan x*is not a direct definition of

*tangent*(defined as a ratio of

*sine*to

*cosine*), but immediately follows from it considering similar triangles Δ

**and Δ**

*ADB***(point**

*ACE***is a base of a perpendicular from point**

*E***to X-axis, not marked on a picture).**

*C*Let's compare the areas of three geometric figures:

(a) triangle Δ

**,**

*ACB*(b) circular sector

**between radiuses**

*ACB***,**

*AC***and arc**

*AB***,**

*CB*(c) triangle Δ

**.**

*ADB*Obviously,

*Area(a) ≤ Area(b) ≤ Area(c)*

Now let's calculate them all.

*Area(a)=AB·sin(x)/2=sin(x)/2*

*Area(b)=πAB²x/(2π)=x/2*

*Area(c)=AB·tan(x)/2=tan(x)/2*

Therefore,

*sin(x) ≤ x ≤ tan(x)*

or, since

*tan(x)=sin(x)/cos(x)*,

*sin(x) ≤ x ≤ sin(x)/cos(x)*

Dividing all by positive

*sin(x)*, we obtain

*1 ≤ x/sin(x) ≤ 1/cos(x)*

The left expression is a constant

*1*, the right expression tends to

*1*as

*x→0*since

*cos(0)=1*. Therefore, by

**squeeze theorem**, the middle expression has a limit as

*x→0*and its limit is

*1*:

*x/sin(x) → 1*as

*x→0*

Therefore,

*sin(x)/x → 1/1=1*as

*x→0*

End of Proof.

*Amazing Limit #2*

**as**

*(1+x)*^{1/x}→ e

*x→0*where

**is a fundamental constant in Calculus, approximately equal to**

*e**2.71*.

*Proof*

First of all, we have to recall the definition of number

**.**

*e*There are many definitions of

**, all equivalent to each other.**

*e*In our course we have chosen the definition of

**as a base of an exponential function**

*e***with**

*y=e*^{x }*steepness*(which is, actually, a trigonometric tangent of a tangential line) at point

**being equal to 1 (that is, the tangential line at**

*x=0***is parallel to a line**

*x=0***).**

*y=x*Using the terminology we are now familiar with, the

*steepness*of a curve that represents the graph of a real function

**or, simply, the**

*y=f(x)**steepness*of a real function

**at some point**

*y=f(x)***is a limit of the ratio**

*x=r***[**as

*f(r+d)−f(r)*]/*d***.**

*d→0*So, we can say that defining characteristic of number

**is**

*e**lim*[(

_{x→0}**)/**

*e*^{x}−1**] =**

*x*

*1*Let's approach this from the position of a function graphs.

The graph of a function

*y = e*^{x}−1

*y = e*^{x}**, its tangent coincides with the graph of function**

*e*

*y = x*Let's analyze now the meaning of the limit we have to prove.

First of all, let's recall that, if

**is a continuous function and**

*f(x)***, then**

*x→r***- this is, actually, a definition of a continuous function.**

*f(x)→f(r)*Next, it can be proven that functions

**and**

*e*^{x}**(a**

*ln(x)**natural logarithm*- a logarithm with a base

**-**

*e***) are continuous.**

*log*_{e}(x)Therefore, instead of proving the limit

**as**

*(1+x)*^{1/x}→ e**,**

*x→0*we can prove that

**as**

*ln(1+x)*^{1/x}→ ln(e)

*x→0*Simple transformation results in the following formulation of our theorem.

Prove that

**as**

*ln(1+x)/x → 1*

*x→0*Let's draw a graph of

**near point**

*ln(1+x)***and compare it with a graph of function**

*x=0***.**

*y=x*Notice that function

**is**

*y=ln(1+x)***inverse**relative to function

**. Here is why:**

*y=e*^{x}−1**⇒**

*y=e*^{x}−1⇒

**⇒**

*y+1=e*^{x}⇒

*ln(y+1)=x*so, an inverse function is

**.**

*y=ln(1+x)*As we know, an inverse function has a graph symmetrical to the original one relative to an angle bisector of the main coordinate angle.

This implies that, if the line

**is tangential to a graph of function**

*y=x***at point**

*y=e*^{x}−1**(which is the origin of coordinates because**

*x=0***as well), the line symmetrical to this tangential line should be tangential to a graph of function**

*y=e*^{0}−1=0**at a point symmetrical to the origin of coordinates.**

*y=ln(x+1)*But the tangential line

**is the angle bisector and is the axis of symmetry, so its symmetrical is itself.**

*y=x*Also, a point symmetrical to the origin of coordinates is itself.

So, we have that the same line

**is tangential to both**

*y=x***and**

*y=e*^{x}−1**at the same point, which proves that**

*y=ln(x+1)***[**

*ln(1+x)−ln(1+0)*]/*x → 1*as

**.**

*x→0*We know that

**. Therefore, the last expression is simplified into**

*ln(1)=0***as**

*ln(1+x)*/*x → 1***.**

*x→0*This completes the analysis of a problem. We are ready now to prove the original theorem.

Let's start from the last statement that we have proven using the considerations of symmetry between the original function

**and its inverse**

*y=e*^{x}−1**, that is:**

*y=ln(1+x)***as**

*ln(1+x)*/*x → 1***.**

*x→0*Let's transform it into an equivalent form using the property of logarithm

*log(a*into the following:

^{b})=b·log(a)**as**

*ln((1+x)*→ 1^{1/x})**.**

*x→0*Using continuity of a function

**, apply it to the above statement:**

*e*^{x}**as**

*(1+x)*^{1/x}→ e^{1}**.**

*x→0*Since

**, we have the original statement proven:**

*e*^{1}=e**as**

*(1+x)*^{1/x}→ e**.**

*x→0*End of proof.

A simple consequence of this limit is that

**as**

*(1+1/n)*^{n}→ e

*n→∞*since

**as**

*1/n → 0*

*n→∞*
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