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Which of these two fractions is greater? ;) watch

1. Anyone got a nifty way to show which of the following is larger?

1) 1/(1+x)

2) (1/x) - 1
2. Subtract (2) from (1) and see if what you get is positive or negative. If it's positive then (1) is greater than (2), if it's negative then vice versa. Although I suspect it might depend on the value of x (but it might not, I haven't tried).
3. you mean which one converges to a larger value? you can show using l'hopital's rule that 1) converges to zero. for the second one, as x tends to infinity 1/x tends to 0 so the equation itself tends to -1. thus 1) tends to a 'larger' value
4. (Original post by Economister)
Anyone got a nifty way to show which of the following is larger?

1) 1/(1+x)

2) (1/x) - 1

Solved it. Where you get it from it's a weird question. You need to set one greater than the other, then use basic rules to deduce something.

Good to consider postive x, then negative x.
5. (Original post by Bjoorn)
you mean which one converges to a larger value? you can show using l'hopital's rule that 1) converges to zero. for the second one, as x tends to infinity 1/x tends to 0 so the equation itself tends to -1. thus 1) tends to a 'larger' value
That is nonsense as you can consider a small neighbourhood around 0, then 1/x will be as large as you want it to be by taking smaller and smaller neighbourhoods around 0.

Also, L'hopital rule can't be used like that.
6. (1/x) - 1 = (1-x)/x

1/(1+x) = (1-x)/(1-x^2)

for x<-1 and x>1, 1-x^2<0 .. the latter fraction is negative if x<-1 and positive if x>1 while the former is always negative given this x domain

for -1<x<1, 1-x^2>0 and the latter is always positive. The former is dependent .. if 0<x<1 then positive otherwise negative for -1<x<0
7. (Original post by Pi-)
(1/x) - 1 = (1-x)/x

1/(1+x) = (1-x)/(1-x^2)

for x<-1 and x>1, 1-x^2<0 .. the latter fraction is negative if x<-1 and positive if x>1 while the former is always negative given this x domain

for -1<x<1, 1-x^2>0 and the latter is always positive. The former is dependent .. if 0<x<1 then positive otherwise negative for -1<x<0
I'm a 12 year old what is this?

Spoiler:
Show
You are doing it wrong. What you said doesn't make sense at all. Seriously, why you introducing power series and long reasoning for something that is just simply solving a quadratic.

OP that isn't what you do.
8. (Original post by Simplicity)
That is nonsense as you can consider a small neighbourhood around 0, then 1/x will be as large as you want it to be by taking smaller and smaller neighbourhoods around 0.

Also, L'hopital rule can't be used like that.
well i just didnt understand the question cos it was worded vaguely so i guessed and explained what i was doing. it's been ages since i did l'hopital and i should have said that but ah well. i'll have a quick look
9. (Original post by Simplicity)
I'm a 12 year old what is this?

Spoiler:
Show
You are doing it wrong. What you said doesn't make sense at all. Seriously, why you introducing power series and long reasoning for something that is just simply solving a quadratic.

OP that isn't what you do.
What are you talking about? I've just written equivalent fractions with a common numerator to those that were given. In essence, what I am saying is that the denominator is the decisive factor as to which fraction is larger.

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