Conditions on L'Hopital's rule (1 Viewer)

leehuan · Apr 22, 2016

I've forgotten why L'Hopitals breaks down for this.

$\lim_{x\rightarrow\infty}{\frac{x+\sin{x}}{x-\sin{x}}}$

Isn't it technically infty/infty

InteGrand · Apr 22, 2016

leehuan said:
I've forgotten why L'Hopitals breaks down for this.

$\lim_{x\rightarrow\infty}{\frac{x+\sin{x}}{x-\sin{x}}}$

Isn't it technically infty/infty

The resulting limit obtained from the L'Hospital step needs to exist (the limit being infinity counting as 'exists' for these purposes). If it doesn't, then the rule doesn't work. In this case, the resulting limit doesn't exist, because it is limit as x -> oo of (1 + cos(x))/(1 - cos(x)), and this function just keeps oscillating as x gets large so no limit exists.

(Of course, the answer to the original limit is 1.)

leehuan · Apr 22, 2016

Oh right, forgot that the existence of the resulting limit is not always assumed to be possible.

But in that case I don't know how to do the question anymore.

$\lim _{ x\rightarrow \infty }{ \frac { 1+\frac { \sin { x } }{ x } }{ 1-\frac { \sin { x } }{ x } } }$

Do I have to determine this one seperately?
$\lim _{ x\rightarrow \infty }{\frac{\sin{x}}{x}}$

InteGrand · Apr 22, 2016

leehuan said:
Oh right, forgot that the existence of the resulting limit is not always assumed to be possible.

But in that case I don't know how to do the question anymore.

$\lim _{ x\rightarrow \infty }{ \frac { 1+\frac { \sin { x } }{ x } }{ 1-\frac { \sin { x } }{ x } } }$

Do I have to determine this one seperately?
$\lim _{ x\rightarrow \infty }{\frac{\sin{x}}{x}}$

That sin(x)/x one can be shown to be 0 via the squeeze theorem.

leehuan · Apr 22, 2016

Alright traditional methods it is. Thanks lol

leehuan · Apr 22, 2016

Got the answer of 1 but I brute forced this one using L'H twice by combining the fractions first. Was wondering if there's a shortcut?

$\lim _{ t\rightarrow 0 }{ \left( \frac { 1 }{ \ln { \left( 1-x \right) } } +\frac { 1 }{ \ln { \left( 1+x \right) } } \right) }$

Paradoxica · Apr 22, 2016

leehuan said:
Got the answer of 1 but I brute forced this one using L'H twice by combining the fractions first. Was wondering if there's a shortcut?

$\lim _{ t\rightarrow 0 }{ \left( \frac { 1 }{ \ln { \left( 1-x \right) } } +\frac { 1 }{ \ln { \left( 1+x \right) } } \right) }$

It can be shown, that for all x>-1

$x \geq \log{(1+x)} \geq \frac{x}{1+x}$

Similarly, for all x<1

$-x \geq \log{(1-x)} \geq \frac{x}{x-1}$

Reciprocate both inequalities and add, to get

$0 \leq \frac{1}{\log{(1+x)}} + \frac{1}{\log{(1-x)}} \leq 2$

But this is useless.

leehuan · Apr 22, 2016

Fair enough. It wasn't hard, just that the product rule felt so dodge.

RealiseNothing · Apr 22, 2016

leehuan said:
Got the answer of 1 but I brute forced this one using L'H twice by combining the fractions first. Was wondering if there's a shortcut?

$\lim _{ t\rightarrow 0 }{ \left( \frac { 1 }{ \ln { \left( 1-x \right) } } +\frac { 1 }{ \ln { \left( 1+x \right) } } \right) }$

Maybe try see if something like this leads anywhere?

$(1-x)^{\frac{1}{\ln(1-x)}+{\frac{1}{\ln(1+x)}}}$

$(1-x)^{\frac{ln(e)}{\ln(1-x)}+{\frac{ln(e)}{\ln(1+x)}}}$

$(1-x)^{log_{1-x}(e)+log_{1+x}(e)}$

$e \times (1-x)^{log_{1+x}(e)}$

leehuan · Apr 22, 2016

RealiseNothing said:
Maybe try see if something like this leads anywhere?

$(1-x)^{\frac{1}{\ln(1-x)}+{\frac{1}{\ln(1+x)}}}$

$(1-x)^{\frac{ln(e)}{\ln(1-x)}+{\frac{ln(e)}{\ln(1+x)}}}$

$(1-x)^{log_{1-x}(e)+log_{1+x}(e)}$

$e \times (1-x)^{log_{1+x}(e)}$

Will investigate tomorrow

Paradoxica · Apr 22, 2016

RealiseNothing said:
Maybe try see if something like this leads anywhere?

$(1-x)^{\frac{1}{\ln(1-x)}+{\frac{1}{\ln(1+x)}}}$

$(1-x)^{\frac{ln(e)}{\ln(1-x)}+{\frac{ln(e)}{\ln(1+x)}}}$

$(1-x)^{log_{1-x}(e)+log_{1+x}(e)}$

$e \times (1-x)^{log_{1+x}(e)}$

I managed to utilise the squeeze theorem, but the original result will require us to take the x^th root of both sides, but you can't take the 0^th root of 1....

seanieg89 · Apr 23, 2016

Notice you are applying a well known function (log) at x close to 1 (where log is quite nice).

A first order Taylor approximation works:

$\log(1+x)=x+O(x^2)$

So

$\frac{\log(1-x^2)}{\log(1-x)\log(1+x)}=\frac{-x^2+O(x^4)}{-x^2+O(x^3)}=\frac{1+O(x^2)}{1+O(x)}\rightarrow 1.$

Don't worry if you don't know what the "big-O" notation stands for, you can still interpret this result by interpreting the first equation as:

$\log(1+x)\sim x \quad \textrm{as }x\rightarrow 0$

in the same way that high schoolers are taught

$\sin(x)\sim x \quad \textrm{as }x\rightarrow 0.$

These type of estimates are really useful for when we have "competing" zeros on the numerator and denominator of a function. All that matters is the "leading order" behaviour of both the numerator and denominator.

Note: It's this kind of discussion of leading order behaviour that can lead one to proving the 0/0 version of L'Hopitals.

leehuan · Jun 9, 2016

$\lim _{ x\to \infty }{ \frac { 1 }{ x } \int _{ x }^{ 4x }{ \cos { \left( \frac { 1 }{ u } \right) } du } }$

How would you do this one?

(I didn't know which thread I should be putting the question under)

InteGrand · Jun 9, 2016

leehuan said:
$\lim _{ x\to \infty }{ \frac { 1 }{ x } \int _{ x }^{ 4x }{ \cos { \left( \frac { 1 }{ u } \right) } du } }$

How would you do this one?

(I didn't know which thread I should be putting the question under)

You can do it via L'Hôpital. To show the numerator goes to infinity, try bounding the integrand.

Paradoxica · Jun 9, 2016

InteGrand said:
You can do it via L'Hôpital. To show the numerator goes to infinity, try bounding and gagging integrand.

ftfy

leehuan · Jun 9, 2016

InteGrand said:
You can do it via L'Hôpital. To show the numerator goes to infinity, try bounding the integrand.

Oops. Thanks lol; I accidentally had the impression that cos(1/x) goes to zero as x is large, not one. Think I just mixed up cos and sin

Though I'm having brainfarts. What function could I apply a comparison test on to properly prove it?

Paradoxica said:
ftfy

r u ok

Paradoxica · Jun 9, 2016

leehuan said:
Oops. Thanks lol; I accidentally had the impression that cos(1/x) goes to zero as x is large, not one. Think I just mixed up cos and sin

Though I'm having brainfarts. What function could I apply a comparison test on to properly prove it?

Recall that

$1-\frac{x^2}{2} < \cos{x} < 1 - \frac{x^2}{2} + \frac{x^4}{24}$

r u ok

No. I'm dying right now... my immune system gave way.

At least I did my X2 task... currently lost fewer marks than anyone else on it... for now.

InteGrand · Jun 9, 2016

There's an easier way, just show that for large enough values of u, cos(1/u) becomes bounded below by some positive constant. From this, show that the integral has to go to infinity.

leehuan · Jun 9, 2016

Yeah I was basically wondering if this was ok:

For some x>M, cos(1/u)>1/2 (or any 0 < x<1) and well obviously the (infinite) integral of a constant diverges by the p-test so that allows a comparison test to be used easily

seanieg89 · Jun 9, 2016

If you need to estimate the integrand anyway, I would just go ahead and use the Taylor bounds you mentioned.

$F(x)=\frac{1}{x}\int_x^{4x} \cos(1/u)\, du\Rightarrow \\ \\ 3=\frac{1}{x}\int_x^{4x}\, du\geq F(x)\geq \frac{1}{x}\int_x^{4x}(1-\frac{1}{2u^2})\, du = 3-\frac{3}{8x^2}\\ \\ \Rightarrow \lim_{x\rightarrow\infty} F(x)=3.$

Conditions on L'Hopital's rule (1 Viewer)

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