L ' Hopital's Rule for x -> infinity

[unscientific]

Homework Statement

Hopital's Rule for x -> ∞ applies the same way as x -> 0.

Homework Equations

As shown in attached pic.

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]

Homework Statement

Hopital's Rule for x -> ∞ applies the same way as x -> 0.

Homework Equations

As shown in attached pic.

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]

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]

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]

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]

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]

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, $\displaystyle h(x)=\frac{1}{g(x)}\quad\to\quad h'(x)=-\frac{g'(x)}{(g(x))^2}\ .$

A similar result holds for k'(x) .

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{h(x)}{k(x)}$

$\displaystyle =\lim_{x\to a}\frac{h'(x)}{k'(x)}$

$\displaystyle =\lim_{x\to a}\frac{g'(x)}{(g(x))^2}\frac{(f(x))^2}{f'(x)}$

$\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}$​

Can you take it from there?

 

[unscientific]

Check that again.

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

A similar result holds for k'(x) .

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{h(x)}{k(x)}$

$\displaystyle =\lim_{x\to a}\frac{h'(x)}{k'(x)}$

$\displaystyle =\lim_{x\to a}\frac{g'(x)}{(g(x))^2}\frac{(f(x))^2}{f'(x)}$

$\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}$​

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]

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]

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]

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\dots$

$\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}$​

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 $\displaystyle \lim_{x\to a}f'(x) \,,\ \lim_{x\to a}g'(x)$ exist & that $\displaystyle \lim_{x\to a}f'(x)\ne0\ .$

Then the following should give the desired result.

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a}g'(x)}{\lim_{x\to a}f'(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}$​

 

[unscientific]

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

Then the following should give the desired result.

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a}g'(x)}{\lim_{x\to a}f'(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}$​

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!