l ' Hopital's Rule for x -> infinity

unscientific

1. The problem statement, all variables and given/known data
Hopital's Rule for x -> ∞ applies the same way as x -> 0.

2. Relevant equations

As shown in attached pic.

3. The attempt at a solution

I tried to prove that x->∞ works the same way as x -> 0, only to get its reciprocal.. Not sure what is wrong with my working..

algebrat

Quote by unscientific

1. The problem statement, all variables and given/known data
Hopital's Rule for x -> ∞ applies the same way as x -> 0.

2. Relevant equations

As shown in attached pic.

3. The attempt at a solution

I tried to prove that x->∞ works the same way as x -> 0, only to get its reciprocal.. Not sure what is wrong with my working..

Why should h or k have a derivative at a? You'll have to come up with different reasoning.

HallsofIvy

You ask about L'Hopital's rule applying "as x-> infinity" but your attachment has x->a while it is f(x) and g(x) that go to infinity. Which are you trying to prove?

Ray Vickson

Quote by HallsofIvy

You ask about L'Hopital's rule applying "as x-> infinity" but your attachment has x->a while it is f(x) and g(x) that go to infinity. Which are you trying to prove?

No only that, you applied Taylor expansions of f(x) and g(x) at x = a, where both functions are infinite! You can't do that.

RGV

unscientific

Quote by HallsofIvy

You ask about L'Hopital's rule applying "as x-> infinity" but your attachment has x->a while it is f(x) and g(x) that go to infinity. Which are you trying to prove?

Sorry for the confusion! I mean f(a) = g(a) = ∞

I earlier showed that l'Hopital's Rule works when its 0/0 but I am now trying to show it works for ∞/∞.

So i took (1/g)/(1/f) to make it 0/0..

But when i use taylor's expansion something's wrong..

SammyS

Quote by unscientific

Sorry for the confusion! I mean f(a) = g(a) = ∞

I earlier showed that l'Hopital's Rule works when its 0/0 but I am now trying to show it works for ∞/∞.

So i took (1/g)/(1/f) to make it 0/0..

But when i use taylor's expansion something's wrong..

What is Taylor's expansion for 1/(f(x)) , expanded about x = a ?

This poses a problem since 1/f(a) is undefined.

Construct two functions, F(x) & G(x) such that F(a) = 0 = G(a), otherwise F(x) = 1/f(x) and G(x) = 1/g(x) .

unscientific

Quote by SammyS

What is Taylor's expansion for 1/(f(x)) , expanded about x = a ?

This poses a problem since 1/f(a) is undefined.

Construct two functions, F(x) & G(x) such that F(a) = 0 = G(a), otherwise F(x) = 1/f(x) and G(x) = 1/g(x) .

Yes i did that, i let 1/g = h, 1/f = k.

then it reduces to [ h'(x)/k'(x) ] which translates to [g'(x)/f'(x)], which is the reciprocal instead..

SammyS

Quote by unscientific

Yes i did that, i let 1/g = h, 1/f = k.

then it reduces to [ h'(x)/k'(x) ] which translates to [g'(x)/f'(x)], which is the reciprocal instead..

Check that again.

So, [itex]\displaystyle h(x)=\frac{1}{g(x)}\quad\to\quad h'(x)=-\frac{g'(x)}{(g(x))^2}\ .[/itex]

A similar result holds for k'(x) .

[itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{h(x)}{k(x)}[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{h'(x)}{k'(x)}[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{(g(x))^2}\frac{(f(x))^2}{f'(x)}[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}[/itex]

Can you take it from there?

unscientific

Quote by SammyS

Check that again.

So, [itex]\displaystyle h(x)=\frac{1}{g(x)}\quad\to\quad h'(x)=-\frac{g'(x)}{(g(x))^2}\ .[/itex]

A similar result holds for k'(x) .

[itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{h(x)}{k(x)}[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{h'(x)}{k'(x)}[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{(g(x))^2}\frac{(f(x))^2}{f'(x)}[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}[/itex]

Can you take it from there?

Yes I ended up with that step. So f(a) = g(a), f/g = 1.

Then we have g'(x)/f'(x), which is the reciprocal, where I am stuck at...

SammyS

Quote by unscientific

Yes I ended up with that step. So f(a) = g(a), f/g = 1.

Then we have g'(x)/f'(x), which is the reciprocal, where I am stuck at...

What is that reciprocal equal to ?

unscientific

Quote by SammyS

What is that reciprocal equal to ?

By right I should get f'(x)/g'(x) as when approximating x→0.

But i'm getting g'(x)/f'(x) instead..

SammyS

Quote by SammyS

[itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\dots[/itex]

[itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}[/itex]

Quote by unscientific

By right I should get f'(x)/g'(x) as when approximating x→0.

But I'm getting g'(x)/f'(x) instead..

Suppose both limits [itex]\displaystyle \lim_{x\to a}f'(x) \,,\ \lim_{x\to a}g'(x)[/itex] exist & that [itex]\displaystyle \lim_{x\to a}f'(x)\ne0\ .[/itex]

Then the following should give the desired result.

[itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=
\frac{\lim_{x\to a}g'(x)}{\lim_{x\to a}f'(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}
[/itex]

unscientific

Quote by SammyS

Suppose both limits [itex]\displaystyle \lim_{x\to a}f'(x) \,,\ \lim_{x\to a}g'(x)[/itex] exist & that [itex]\displaystyle \lim_{x\to a}f'(x)\ne0\ .[/itex]

Then the following should give the desired result.

[itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=
\frac{\lim_{x\to a}g'(x)}{\lim_{x\to a}f'(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}
[/itex]

I see! then the f(x)/g(x) cancels on both sides leaving only one f(x)/g(x) on the right, then you bring over the g'(x)/f'(x) to the left to give f'(x)/g'(x) as desired!

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HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	You ask about L'Hopital's rule applying "as x-> infinity" but your attachment has x->a while it is f(x) and g(x) that go to infinity. Which are you trying to prove?