Limit of a trigonometric function

[BvU]

Glad you take it like a good sport. My point is that I carefully try to seduce you to build up some self-confidence in the things you already are quite able to do and check by yourself (just look at the careful formulation !). A bit of tenacity and stubbornness can come in quite useful, and you can always go back and check steps later on.

 

[SammyS]

is step 32 correct?

It looks like you need to do at least one more step.

Added in Edit:
Well, if you're correct in post #32, that one more step also gives the form 0/0 .

A step after that should give you something good.

Added in 2nd Edit:

In the expression for g''(x) in post #32, the 96 should be 48 instead.

Added in 3rd Edit:

Have you tried Taylor expansions of numerator & denominator ?

 

[chwala]

agreed seen the mistake $g''(x)=-16(sin 2x)^4+ 8cos 2x.(3(sin 2x)^2.2 cos 2x)$ → $g''(x)=-16(sin 2x)^4+48(sin2x)^2.cos 2x$

 

[chwala]

now $f'''(x)= -9 sin x +27 sin 3x$
$f'''(x) = 27 sin 3x - 9 sin x$
$g'''(x)= -64(sin 2x)^3.2cos 2x+ 192sin 2x.(cos 2x)^2+48(sin 2x)^2.-2sin 2x$
$g'''(x)= -128 (sin 2x)^3.cos 2x+192sin 2x(cos 2x)^2-96(sin 2x)^3$
again here we get the indeterminate form 0/0
$f''''(x)=81 cos 3x-9cos x$
$g''''(x)= 256(sin 2x)^4-384(sin 2x)^2.cos 2x+384(cos 2x)^3-384(sin 2x)^2cos 2x-576(sin 2x)^2cos 2x$
$g''''(x)=256(sin 2x)^4-384(sin 2x)^2.cos 2x+384(cos 2x)^3-960(sin 2x)^2.cos 2x$
$g''''(x)=256(sin 2x)^4-1344(sin 2x)^2cos 2x$
is this step correct...do we reduce $sin 2x=2 sin x cos x$?
and $cos 3x=cos(x+2x)=cos x.cos2x-sinx.sin2x$
are these steps necessary, let's try we close this...long since i looked at Taylor series...

 

[BvU]

I recognize $f''''$. It no longer gives zero but 72 for $x \downarrow 0$.
The numerator also gives nonzero for $x \downarrow 0$, but there I expect 384 (just a hunch ) so there is something to be improved.

Reducing isn't desired at all here, just taking the limit is enough once the numerator gives nonzero.

 

[ehild]

Both the numerator and the denominator can be expressed as polynomials of cos(x) and then applying L'Hospital is much easier.

 

[BvU]

agreed seen the mistake $g''(x)=-16(sin 2x)^4+ 8cos 2x.(3(sin 2x)^2.2 cos 2x)$ → $g''(x)=-16(sin 2x)^4+48(sin2x)^2.cos 2x$

Poor Chwala. Lost in ciphering.$$
g''(x)=-16(sin 2x)^4+ 8cos 2x.(3(sin 2x)^2.2 cos 2x)\rightarrow \\ g''(x)=-16 \sin^4 2x+48 \sin^2 2x \cos^2 2x$$ $\quad$ The last $\cos 2x$ needs to be squared. The easy check is that the sum of powers is and stays 4.

[edit] you still had that square in post #32 so I'll consider it as a sloppy typo. Painful, becasue you build on it afterwards.

(PS you want to use \cos and \sin =instead of cos and sin in $\TeX$)​

If you don't like spoilers, don't read the remainder of this post )

Here's the helping hand so you can concentrate on thinking:

$$ g'''(x)= -128 (sin 2x)^3.cos 2x+192sin 2x(cos 2x)^2-96(sin 2x)^3 $$

now becomes
$$g'''(x)= -128 \sin^3 2x \cos 2x + 192 \sin 2x \cos^3 2x - 192 \sin^3 2x \cos 2x $$

note the cosine has a power 3, not 2, in the middle term​

and here my natural laziness takes over: I have already seen that $\ f''''\$ goes to nonzero for $\ x\downarrow 0\$ so I am not interested in terms of $\ g''''\$ that do go to zero -- the ones with sines in them. That leaves $$ g''''(x)= 384 \cos^4 2x + \text {sine terms}$$ (really happy with the 384 )

At this point we look back and check what we've done. If so desired, a numerical check on the calculator or in excel ( x f g f/g, respectively):

0.1000 0.0002990016233728 0.0015578416669531 0.191933
0.0100 0.0000000299989995 0.0000001599573385 0.187544
0.0010 0.0000000000029995 0.0000000000160000 0.187469
0.0001 0.0000000000000000 0.0000000000000016 0.000000

(I have big problems getting equal spacing fonts in PF that respect spacing and tabbing )

As you see, f and g race to zero very rapidly -- which is in line with our having to differentiate four times. And 72/384 = 0.1875 so that looks fine !

----

If you are all happy and content with this result, we are going to look at what I dangled in front of you in post #33 (and Sammy mentioned it in post #37) and what Elizabeth brought in in post #41.

 

[chwala]

lol...i must have been tired oooh yeah,i left out $384(cos2x)^3$ limit becomes ${72/384}$

 

[chwala]

I recognize $f''''$. It no longer gives zero but 72 for $x \downarrow 0$.
The numerator also gives nonzero for $x \downarrow 0$, but there I expect 384 (just a hunch ) so there is something to be improved.

Reducing isn't desired at all here, just taking the limit is enough once the numerator gives nonzero.

small error you meant denominator not numerator...

 

[BvU]

small error you meant denominator not numerator...

Correct. Sorry.

Now, what about his Taylor series approach ? As you see from the description, it's a way to write a function as a polynomial where the derivatives appear in the coefficients. So instead of differentiating numerator and denominator until one of the two is nonzero in the limit, you look at these coefficients.

Note: I don't think I ever heard of l'Hôpital until I started to 'work' for PF.​

In the toolkit of nearly all scientists there is the first (or sometimes a few) of these Taylor coefficients for frequently occurring functions. They are extremely useful for estimating behaviour af composite functions, for taking limits, for estimation, numerical procedures etc, etc. In this exercise we need

​

​

(pictures from Wikipedia - having a hard time letting them appear as pictures)​

I had to look them up, for a reason:

'Everyone' knows that 'sine goes like x' and that 'cosine goes like $\ 1-x^2/2\$ for small x. But your slightly sadistic exercise composer carefully concocted a quotient that goes to zero like $\ {ax^4\over b x^4}\$.

In the denominator you see a $\ 2x^4\$ as the first nonzero term in the Taylor series. So in this exercise the denominator goes like $\ 16 x^4 \$And that's all you need to know for this exercise (that sure beats differentiating four times and derailing once or twice on the way, doesn't it ?). If the numerator goes like 'x to the less than four' the limit doesn't exist and for 'x to the more than four' the limit is zero.

And in the numerator some work is required, for the first coefficients are 'coincidentally' canceling: 9-8+1 . And so is the next coefficient (for $\ x^2\ : \ 9 * {1\over 2} - {1\over 2} * 9 \$) so we need terms up to $x^4$ (of course -- look at the denominator).

And even lazy me was curious enough to work on $$ - 9\left (1-{x^2\over 2} + {x^4\over 24}\right ) + \left (1-{(3x)^2\over 2} + {(3x)^4\over 24}\right )$$ until I had $\ {72\over 24 }\$. (That's why I expected 384 = 16 * 24 in your final denominator fourth derivative). Granted, it's work, but a lot less work than differentiating four times and derailing once or twice on the way.

- - - -

So much for the lecture notes. I called this a quasi-alternative approach because you still work with the derivatives that are tucked away in the coefficients of the Taylor series.

@ehild came (#41) with the suggestion to express numerator and denominator as polynomials of cosines. I still see a lot of work coming at me to then do l'Hôpital but perhaps she saw a shorter path through ?

 

[ehild]

Note: I don't think I ever heard of l'Hôpital until I started to 'work' for PF.​

At this part of the world, l'Hospital rule is taught first, as a method to find the limit of functions f(x)/g(x) when the limit of both f and g are zero. Taylor and Maclauren series come later.
​

@ehild came (#41) with the suggestion to express numerator and denominator as polynomials of cosines. I still see a lot of work coming at me to then do l'Hôpital but perhaps she saw a shorter path through ?

Yes, the Taylor series method is easier, if somebody is familiar with it.
Transforming the numerator and denominator to polynomials of cos(x) requires high-school trigonometry only.
The result is $$\frac{1}{4}\left(\frac{2-3y+y^3}{y^4-2y^6+y^8}\right)$$ where y=cos(x), and we have to take the limit when y-->1. It needs l'Hopital twice.

 

[Mark44]

At this part of the world, l'Hospital rule is taught first, as a method to find the limit of functions f(x)/g(x) when the limit of both f and g are zero. Taylor and Maclauren series come later.

Same here in my part of the world (U.S.).

 

[BvU]

Dutch are pragmatic.

 

[SammyS]

...

@ehild came (#41) with the suggestion to express numerator and denominator as polynomials of cosines. I still see a lot of work coming at me to then do l'Hôpital but perhaps she saw a shorter path through ?

Just need to get cos(3x) in terms of cos(x):

$\cos(3x)=4\cos^3(x)-3\cos(x)$​

The derivative of the numerator is then $\ 12\sin^3(x)\ .$

The derivative of the denominator then has a factor of $\ \sin^3(2x)\$ which is easily reduced.

$\sin^3(x)\$ in both numerator & denominator cancel. The answer is readily at hand.

+ + + + for @ehild !

 

[ehild]

Dutch are pragmatic.

Or you just missed the lecture about l'Hospital rule ??

 

[ehild]

Just need to get cos(3x) in terms of cos(x):

$\cos(3x)=4\cos^3(x)-3\cos(x)$​

The derivative of the numerator is then $\ 12\sin^3(x)\ .$

The derivative of the denominator then has a factor of $\ \sin^3(2x)\$ which is easily reduced.

$\sin^3(x)\$ in both numerator & denominator cancel. The answer is readily at hand.

Sammy is the greatest!

 

[SammyS]

As a followup to post #49, which showed that if you follow @ehild's advice, it is only necessary to apply L'Hôpital's rule one time:

This implies that simplifying OP's result (once corrected) after the first application of L'Hôpital's rule should give the desired result with no further need to appeal to L'Hôpital .

That is indeed the case.

 

[BvU]

All extra-smartness aside, I pragmatically stick to the approach in #45 or this kind of exercises

 

[SammyS]

All extra-smartness aside, I pragmatically stick to the approach in #45 or this kind of exercises

I agree that using the Taylor series approach can give the limit more quickly and with less chance for error. It's what I used to get my initial answer.

(Apparently I have some Dutch heritage.)

 

[chwala]

Good read am learning.