Revision for exams. Unsure of problems

RolandIRL

hi all,
doing a bit of revision for maths, and i'm having trouble trying to prove that

lim sin x/x = 1 as x tends to 0

i tried to prove it using the sandwich lemma
(< means less than or equal to in this post, for simplicity's sake)
-1/x < sin x/x < 1/x

taking the limit as x -> 0 , i'm getting that -(infinity) < lim sinx/x < (infinity)
so what's the most straightforward way to prove the limit?

i'm also having difficulty proving the triangle inequality
|a + b| < |a| + |b|

i'm going round in circles, trying to prove this.

thanks for helping
whiteman19

Fremen

The limit for sin(x)/x is found via l'hopital's rule. It should be in any calculus textbook. That, or you could expand Sin in a power series around x=0, divide by x and take a limit.

For the triangle inequality, split it up into four cases: (a<0, b<0), (a<0,b>0), (a>0,b<0), (a>0,b>0).

RolandIRL

thanks Fremen, turns out i didn't actually need it after all. it didn't come up in the exam today. :rolleyes:

Fremen

Not that I'm trying to sound like your mum, but just because it doesn't come up in the exam doesn't mean you don't nead it. The triangle inequality and l'Hopital are of fairly fundamental importance. If you don't understand them now, you'll be screwed next year.

RolandIRL

if i'm doing maths....undecided as of yet. but definitely doing maths physics or maths. not sure which to pick.
i will learn it next year if i have to, just not now when i've other exams to study for, which i'm supposed to be doing now, but the lure of boards was too strong :P

equivariant

Fremen wrote: »

The limit for sin(x)/x is found via l'hopital's rule. It should be in any calculus textbook. That, or you could expand Sin in a power series around x=0, divide by x and take a limit.

For the triangle inequality, split it up into four cases: (a<0, b<0), (a<0,b>0), (a>0,b<0), (a>0,b>0).

Using l'hopital's rule for the limit of sin(x)/x is circular because the (standard) calculation of the derivative of sin requires the calculation of this limit.

sponsoredwalk

For sinx/x there are two intuitive ways to remember it is 1.

sinx ≈ x for all x near 0 i.e. for really really small x you can say that;

x = sinx

so, we multiply both sides by (1/x)

1 = sinx/x

Also, by the fact that sin(-x) = - sinx;

1 = sin(-x)/(-x) = -(-sinx)/x = sinx/x it is symmetric.

The proof is done here: http://www.youtube.com/watch?v=Ve99biD1KtA

but if you know that;

sinx < x < tanx (∀ 0 < x < π/2 )

, you use the odd property of sinx & tanx and you follow the same algebraic argument as in the video you'll get the proof without drawing the picture.

That's the best I have on this little theorem anyway

gerardduff

equivariant wrote: »

Using l'hopital's rule for the limit of sin(x)/x is circular because the (standard) calculation of the derivative of sin requires the calculation of this limit.

Not sure about that. Here's a calculation of d/dx [sin(x)] which appeals to infinite series/perturbation theory but is not an approximation as the limit takes over towards the end of the calculation. We just need to set up the calculation so the limit doesn't take us for mugs. I think it's fairly standard. First of all, write down a few standard formuae,

[latex]\dispaystyle \sin(A + =\sin A \cos B +\cos A \sin B[/latex]

Also,

[latex]\dispaystyle sin\theta=\theta - \frac{1}{3!}\theta^3 + \frac{1}{5!}\theta^5 - \frac{1}{7!}\theta^7 + ...[/latex]


and,

[latex]\dispaystyle cos\theta=1-\frac{1}{2!}\theta^2 + \frac{1}{4!}\theta^4 - \frac{1}{6!}\theta^6 +...[/latex]

Now, we need to evaluate

[latex]\dispaystyle Lim_{\Delta x-> 0}\frac{\sin(x + \Delta x) - \sin x}{\Delta x}[/latex]

which becomes, after some manipulations involving the first of the three formula,

[latex]\dispaystyle Lim_{\Delta x-> 0}\frac{\sin x\cos\Delta x+\cos x\sin\Delta x-\sin x}{\Delta x}[/latex]

Now replace [latex]\dispaystyle \cos\Delta x[/latex] and [latex]\sin\Delta x[/latex] with their infinite series. In the numerator the [latex]sin x[/latex] terms cancel. All other terms contain a factor in [latex]\Delta x[/latex], so they will cancel with the denominator. Now, apply the limit and the result falls out.

The calculation in the original post can be done the same/similar way.

MathsManiac

But equivariant did refer to the "standard" derivation. Using infinite series is a bit of an overkill for such an elementary result.

Referring to the original post, it is worth pointing out that you can indeed do this with the sandwich lemma.

Draw a sector of a circle of angle x < Pi/2 and of radius 1.

Join the endpoints of the arc to inscribe a triangle.

Also, erect a perpendicular to one radius at the endpoint of the arc until this perpendicular crosses the other radius extended.

The area of the sector is between the area of the inscribed triangle and the outer triangle.

This gives sin x < x < tan x

Divide across by sin x to get 1 < x/(sin x) < sec x.

Invert if you like to get 1 > (sin x) / x > cos x.

Apply the limit and Bob's your uncle.

gerardduff

MathsManiac wrote: »

Invert if you like to get 1 > (sin x) / x > cos x.
Apply the limit and Bob's your uncle.

Ok, very good. This is a neat geometrical method/approach which some people will appreciate more than others.

[latex]1>\frac{\sin x}{x}>1[/latex] then [latex]\frac{\sin x}{x}=1 [/latex].

I don't know what the so called standard derivation is but I'd lay money it ain't your one.

In my book the method I outlined should be standard in the sense of being in the repetoire of anyone doing leaving cert honours maths or studying maths/physics/engineering at third level. The limit in the original post and indeed the whole idea of a limit occurs in calculus and therefore it is natural to use a method which should appear in any calculus text.

Again, may I say, I like your geometrical approach, just in case you think otherwise. In my opinion, methods such as yours are made all the more enjoyable when the calculus proof is learned and understood. The bottom line is that the method I used is more instructive to a typical student of physics/mathematics, at least that's my experience.

MathsManiac

When equivariant referred to the standard derivation of the derivative of sin(x), I assume he meant the one usually done at leaving cert, and which is also common in elementary calculus texts.

It involves expressing sin(x+h) - sin(x) as a product: 2cos(x + h/2)sin(h/2).

Then, in taking the limit as h -> 0, you then use the fact that the limit of (sin(h/2))/(h/2) is 1, (having established this limit beforehand).

It's my understanding that this is what underlay the discomfort expressed at then using using l'Hopital's rule to establish the limit concerned.

I accept that your derivation is fully correct. However, it does rely on properties of infinite series that are not trivial to establish, and which are generally not encountered before elementary differential calculus.

Defining sin and cos by their power series is of course a useful way to define them, since it allows them to be extended into the realm of complex numbers, but I think it's fair to say that most people encounter these first as trigonometric ratios, and then as co-ordinates on the unit circle, and do a lot of work with them before they move on to infinite series.

equivariant

MathsManiac wrote: »

Using infinite series is a bit of an overkill for such an elementary result.

Indeed it is. However this result is always a tricky point of the basic calculus curriculum. I don't believe that there is any satisfactory way to show that

[latex] \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1 [/latex]

using only the standard 1st year university calculus curriculum.

There are two approaches to take, depending on how you define sin. The easiest way is to define sin using the standard power series expansion. Then it is relatively easy to evaluate the limit above using basic facts about series. However, the problem with that is that it is quite difficult to show that the power series definition of sin agrees with the elementary definition given in terms of ratios of side lengths in a triangle.

The other option is to start with the standard definition of sin x as the y-coordinate of the point p on the unit circle such that the arc length from (1,0) to p is x. However, to calculate the limit above, one must then treat the concept of arclength rigourously (which requires a fair bit of integral calculus) in order to get the bounds one needs to apply (for example) the sandwich lemma.

In either approach there are difficulties. I think, for this reason, the proof of this limit calculation is fudged in most basic calculus courses.

If anyone knows another approach, I'd love to see it.

PS Basically, my point is that this limit is deceptive. It is not really an elementary result (even though it is intuitively plausible), but it is required in order to make use of trig functions in standard calculus courses.

sponsoredwalk

Equivariant, I wonder what you'd find wrong or "fudged" with the following explanation of this limit. I thought I had it down but I may have assumed something stupid.

1: ∀ 0 < Θ < (pi)/2 ⇒ sinΘ < Θ < tanΘ

from the unit circle definition.

2: by virtue of the "oddness" of the sine function, i.e. sin(-Θ) = - sin(Θ) we can see that the inequality sinΘ < Θ < tanΘ will be symmetric through the origin. We can use the absolute value's and implement the squeeze/sandwich theorem.

3: Multiplying the inequality by |1/sinΘ|

[latex] |sin \theta | \cdot | \frac{1}{sin \theta } | < |\theta | \cdot | \frac{1}{sin \theta } |< |\frac{sin \theta }{cos \theta } | \cdot | \frac{1}{sin \theta } | [/latex]

[latex] 1 < | \frac{\theta }{sin \theta } |< |\frac{1}{cos \theta } | [/latex]

4: Take limits:

[latex] \lim_{\theta \to 0} 1 < \lim_{\theta \to 0} | \frac{\theta }{sin \theta } |< \lim_{\theta \to 0} |\frac{1}{cos \theta } | [/latex]
Taking the rightmost limit with cosΘ we know that cos(0) = 1 and by the squeeze principle we see that the function sinΘ/Θ is bounded between 1 & 1.

(this verifies the fact that the limit sinx/x can be inverted to produce the same result!)

equivariant

sponsoredwalk wrote: »

Equivariant, I wonder what you'd find wrong or "fudged" with the following explanation of this limit. I thought I had it down but I may have assumed something stupid.

1: ∀ 0 < Θ < (pi)/2 ⇒ sinΘ < Θ < tanΘ

from the unit circle definition.

2: by virtue of the "oddness" of the sine function, i.e. sin(-Θ) = - sin(Θ) we can see that the inequality sinΘ < Θ < tanΘ will be symmetric through the origin. We can use the absolute value's and implement the squeeze/sandwich theorem.

3: Multiplying the inequality by |1/sinΘ|

[latex] |sin \theta | \cdot | \frac{1}{sin \theta } | < |\theta | \cdot | \frac{1}{sin \theta } |< |\frac{sin \theta }{cos \theta } | \cdot | \frac{1}{sin \theta } | [/latex]

[latex] 1 < | \frac{\theta }{sin \theta } |< |\frac{1}{cos \theta } | [/latex]

4: Take limits:

[latex] \lim_{\theta \to 0} 1 < \lim_{\theta \to 0} | \frac{\theta }{sin \theta } |< \lim_{\theta \to 0} |\frac{1}{cos \theta } | [/latex]
Taking the rightmost limit with cosΘ we know that cos(0) = 1 and by the squeeze principle we see that the function sinΘ/Θ is bounded between 1 & 1.

(this verifies the fact that the limit sinx/x can be inverted to produce the same result!)

The first line. If you use the unit circle definition then how do you obtain those bounds? The problem with the 'unit circle definition' of sin or cos is that it hides a lot machinery under the bonnet. The unit circle definition requires a rigourous notion of arc length along a circle, which already requires a fair amount of integral calculus. Even if the definition is set up properly. to obtain the inequalities sinΘ < Θ < tanΘ is not trivial.

sponsoredwalk

You got me

I spent 10 minutes trying to figure out the way to show that.

In Apostol's Calculus he takes this as one of his "four fundamental properties", or "axioms" of trigonometry.

Also, you can intuit why sinx < x just by looking at the sine curve and realising x grows faster than y = sinx but I'll admit I cannot prove this yet

Fremen

sponsoredwalk wrote: »

Also, you can intuit why sinx < x just by looking at the sine curve and realising x grows faster than y = sinx but I'll admit I cannot prove this yet

It should be fairly straightforward. We're working in radian measure, and the radius in this case is 1. That means theta is the arc length from the point (1,0) to (Cos theta, Sin theta). On the other hand, Sin theta is the length of the perpendicular dropped from (Cos theta, Sin theta) to the X axis.

To summarise: curvy line is longer than straight line.

MathsManiac

equivariant wrote: »

Indeed it is. However this result is always a tricky point of the basic calculus curriculum. I don't believe that there is any satisfactory way to show that

[latex] \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1 [/latex]

using only the standard 1st year university calculus curriculum.

There are two approaches to take, depending on how you define sin. The easiest way is to define sin using the standard power series expansion. Then it is relatively easy to evaluate the limit above using basic facts about series. However, the problem with that is that it is quite difficult to show that the power series definition of sin agrees with the elementary definition given in terms of ratios of side lengths in a triangle.

The other option is to start with the standard definition of sin x as the y-coordinate of the point p on the unit circle such that the arc length from (1,0) to p is x. However, to calculate the limit above, one must then treat the concept of arclength rigourously (which requires a fair bit of integral calculus) in order to get the bounds one needs to apply (for example) the sandwich lemma.

In either approach there are difficulties. I think, for this reason, the proof of this limit calculation is fudged in most basic calculus courses.

If anyone knows another approach, I'd love to see it.

PS Basically, my point is that this limit is deceptive. It is not really an elementary result (even though it is intuitively plausible), but it is required in order to make use of trig functions in standard calculus courses.

I appreciate the point you're making, but this issue is not insurmountable. A rigorous treatment of trigonometry based on a rigorous treatment of geometry is entirely possible. In, for example, "Geometry with Trigonometry" by Patrick D. Barry (2001, Horwood Publishing), this limit is properly established without reference to arc length. The final stage is reminiscent of to the argument I outlined above, except that their is no explicit reference to the area of the sector. (There is, in its place, a properly established function of the angle measure x. This function turns out to be, in essence, the area of the sector, without claiming to be!)

equivariant

MathsManiac wrote: »

I appreciate the point you're making, but this issue is not insurmountable. A rigorous treatment of trigonometry based on a rigorous treatment of geometry is entirely possible. In, for example, "Geometry with Trigonometry" by Patrick D. Barry (2001, Horwood Publishing), this limit is properly established without reference to arc length. The final stage is reminiscent of to the argument I outlined above, except that their is no explicit reference to the area of the sector. (There is, in its place, a properly established function of the angle measure x. This function turns out to be, in essence, the area of the sector, without claiming to be!)

hmm. I don't have a copy of Paddy Barry's book at home. However, I wonder how does he deal with angle measure without referring to arc length?

More precisely, I can see that one could define sin to be a function of angles (rather than real numbers). But in order to make sense of an expression like [latex] \lim_{x\rightarrow 0} \frac{sin x}{x}[/latex]
one must have some way of mapping real numbers to angles, which seems to entail a notion of arc length?

PS For example here is an apparently rigourous proof of the limit calculation - http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx

However, I think that this proof assumes at the outset that we can identify real numbers with angles and moreover compares arc lengths to lengths of line segments without rigourous justification (essentially he just appeals to a picture). Is Barry's proof fundamentally different to this one?

MathsManiac

I agree that the derivation on that link assumes that there is such a thing as the length of an arc and that arc-length has certain properties, and that this treatment therefore fails to address the issues you raised.

I haven't time to outline Barry's treatment now; I'll try to get a chance this evening.

MathsManiac

MathsManiac wrote: »

I haven't time to outline Barry's treatment now; I'll try to get a chance this evening.

Ok. whistle-stop tour:

An "angle support" is defined as two half-lines with the same initial point. This gives (except in the case of a straight angle, which is handled seperately) an interior and exterior region. (The interior region of ABC is the intersection of the closed halfplane with edge AB and containing C with the closed halfplane with edge AC and containing B. The exterior region is defined as the union of the two other closed half planes.) An angle is defined as a couple consisting of an angle support and a corresponding interior or exterior region. (If it's an interior region, it's referred to as a wedge-angle; if it's an exterior region, it's a reflex angle.)

Distance and angle measure are taken as primitive concepts defined axiomatically. Distance is a function from PI X PI to R, (where PI is the plane) with basic axiomatic properties like the fact that if C lies on the segment AB, then |AB|=|AC|+|CB|.

Similarly, degree measure is taken as a function from the set of all wedge and straight angles to R. (Codomain is R but range is the closed interval [0,180].) The axiom governing it is comparable to the one for distance. The choice of 180 as the measure of a straight angle in the axiom is arbitrary, but obviously yields the usual meaning of "degrees". The axiom has things like: if the half-line AD is in the interior region of angle BAC, then the measure of BAC is the sum of the measures of BAD and DAC. Later, degree measure is extended to reflex angles.

There's no reference to curves of any sort in this, and circles haven't been defined yet.

A couple of chapters later, circles are defined, and you get to deal with chords, tangents and all that jazz.

A couple of more chapters later, cosine and sine of an angle are introduced. You basically construct a circle of any radius, centred at the vertex of the angle. At the point where one of the arms intersects the circle, you drop a perpendicular to the line containing the other arm. This allows the cosine and sine of an angle to be defined, but there's a bit of work to deal with the various cases in the different quadrants properly.

Note that cosine and sine as defined are functions of angles themselves, not of angle-measures.

By this time also, the idea of a frame of reference and cartesian co-ordinates have been covered, and now the idea of an angle in standard position with respect to a frame of reference is introduced. Definition of addition and subtraction of angles in standard position is covered, and the sine and cosine rule are done. He also proves that if two angles are equal in measure, then their sines are equal and their cosines are equal.

A few chapters later, the idea of the cosine and sine as functions of numbers (as opposed to angles) is introduced. To avoid confusion between these distinct meanings, Barry uses "c" and "s" for these new functions. The function "c(x)", whose domain is the real interval [0,360] is basically defined as the cosine of an angle whose degree measure is x, and similarly s(x). These are well defined by the previously covered material.

Now we're getting close to establishing the limit that started this whole discussion!

Given any x in [0,360], Barry defines two sequences un(x) and vn(x). un is basically the total area of 2^(n+1) triangles inscribed in the arc (or circle) of radius 1 like pizza slices, and vn is the total area of a comparable set of triangles "outside" (i.e. with a side tangential to the arc). He proves that un is non-decreasing and vn is non-increasing, that un is always less than vn, and that they tend to a commom limit mu(x) that lies between un and vn for all n. He then proves that there is a real number p such that mu(x) = px.

This function mu(x) is essentially the area of a sector of a circle with radius 1 and angle x degrees, but he doesn't define this to be the area - it is simply a function of x with certain proven properties.

He defines Pi to be mu(360).

Using the same un and vn sequences, and similar arguments to those used in establishing the function mu, he shows that the limit of s(t)/t as t goes to zero (from above) is Pi/180.

The limit from below is easily handled, and then the derivative of s and c are established using that limit. The derivatives are s'(x) = (Pi/180)*c(x) and c'(x) = -(Pi/180)*s(x).

Radian measure of an angle is now defined as Pi/180 times the degree measure, and radian versions of the c and s functions are defined.

With these techniques, as Barry points out, it is clearly possible to go on and define area of a disk or a sector, and to define the length of an arc of a circle. However, for speed and convenience, (since he's at the end of the book!) he just gives definitions based on integrals. The point is, however, that the limit we were interested in is established before arc length or area of a sector are defined (even though they are ready to be defined).

So there you go. If you've any questions, read the book!

equivariant

MathsManiac wrote: »

Ok. whistle-stop tour:

An "angle support" is defined as two half-lines with the same initial point. This gives (except in the case of a straight angle, which is handled seperately) an interior and exterior region. (The interior region of ABC is the intersection of the closed halfplane with edge AB and containing C with the closed halfplane with edge AC and containing B. The exterior region is defined as the union of the two other closed half planes.) An angle is defined as a couple consisting of an angle support and a corresponding interior or exterior region. (If it's an interior region, it's referred to as a wedge-angle; if it's an exterior region, it's a reflex angle.)

Distance and angle measure are taken as primitive concepts defined axiomatically. Distance is a function from PI X PI to R, (where PI is the plane) with basic axiomatic properties like the fact that if C lies on the segment AB, then |AB|=|AC|+|CB|.

Similarly, degree measure is taken as a function from the set of all wedge and straight angles to R. (Codomain is R but range is the closed interval [0,180].) The axiom governing it is comparable to the one for distance. The choice of 180 as the measure of a straight angle in the axiom is arbitrary, but obviously yields the usual meaning of "degrees". The axiom has things like: if the half-line AD is in the interior region of angle BAC, then the measure of BAC is the sum of the measures of BAD and DAC. Later, degree measure is extended to reflex angles.

There's no reference to curves of any sort in this, and circles haven't been defined yet.

A couple of chapters later, circles are defined, and you get to deal with chords, tangents and all that jazz.

A couple of more chapters later, cosine and sine of an angle are introduced. You basically construct a circle of any radius, centred at the vertex of the angle. At the point where one of the arms intersects the circle, you drop a perpendicular to the line containing the other arm. This allows the cosine and sine of an angle to be defined, but there's a bit of work to deal with the various cases in the different quadrants properly.

Note that cosine and sine as defined are functions of angles themselves, not of angle-measures.

By this time also, the idea of a frame of reference and cartesian co-ordinates have been covered, and now the idea of an angle in standard position with respect to a frame of reference is introduced. Definition of addition and subtraction of angles in standard position is covered, and the sine and cosine rule are done. He also proves that if two angles are equal in measure, then their sines are equal and their cosines are equal.

A few chapters later, the idea of the cosine and sine as functions of numbers (as opposed to angles) is introduced. To avoid confusion between these distinct meanings, Barry uses "c" and "s" for these new functions. The function "c(x)", whose domain is the real interval [0,360] is basically defined as the cosine of an angle whose degree measure is x, and similarly s(x). These are well defined by the previously covered material.

Now we're getting close to establishing the limit that started this whole discussion!

Given any x in [0,360], Barry defines two sequences un(x) and vn(x). un is basically the total area of 2^(n+1) triangles inscribed in the arc (or circle) of radius 1 like pizza slices, and vn is the total area of a comparable set of triangles "outside" (i.e. with a side tangential to the arc). He proves that un is non-decreasing and vn is non-increasing, that un is always less than vn, and that they tend to a commom limit mu(x) that lies between un and vn for all n. He then proves that there is a real number p such that mu(x) = px.

This function mu(x) is essentially the area of a sector of a circle with radius 1 and angle x degrees, but he doesn't define this to be the area - it is simply a function of x with certain proven properties.

He defines Pi to be mu(360).

Using the same un and vn sequences, and similar arguments to those used in establishing the function mu, he shows that the limit of s(t)/t as t goes to zero (from above) is Pi/180.

The limit from below is easily handled, and then the derivative of s and c are established using that limit. The derivatives are s'(x) = (Pi/180)*c(x) and c'(x) = -(Pi/180)*s(x).

Radian measure of an angle is now defined as Pi/180 times the degree measure, and radian versions of the c and s functions are defined.

With these techniques, as Barry points out, it is clearly possible to go on and define area of a disk or a sector, and to define the length of an arc of a circle. However, for speed and convenience, (since he's at the end of the book!) he just gives definitions based on integrals. The point is, however, that the limit we were interested in is established before arc length or area of a sector are defined (even though they are ready to be defined).

So there you go. If you've any questions, read the book!

Thanks for that. I basically see the idea. I guess that he essentially uses some integral calculus (u_n and v_n are basically Riemann sums, if I understand correctly). But because he is only dealing with a very specific situation, he doesn't need the full theory of the Riemann integral.

Anyway, I see now that he has a fully rigourous proof of this limit. On the other hand it would probably be impractical to cover all this stuff in a first university analysis course.