Wednesday, May 14, 2008

The number 1 does not exist.

Edit: I apologize for the crappy formatting. I'm having some trouble with blogger and using small fonts. I'll probably go harass Abbie or someone else who knows more about blogger's interface.

I'm a bit of a fan of fallacious proofs. This will be the first in a series of a posts over the next few weeks of some of my favorite fallacious proofs. Today we are going to prove that 1 does not exist.

Now you may recall from calculus that L'Hospital's rule states that if I have a limit of the form f/g and both f and g are both going to infinity or are both going to 0 then the limit is the same as the limit of f'/g' where f' is the derivative of f and g' is the derivative of g.

So for example, if I
had lim x ->oo (x^2+1)/(2x^2) as x goes to infinity this would be equal to lim x ->oo 2x/4x=1/2 since the derivative of x^2+1 is 2x and the derivative of 2x^2 is 4x and both the top and the bottom of the original limit are going to infinity.

Another example would be if we had lim
x ->oo (x+sin x)/x^2. We set f(x)= x+sin x and g(x)=x^2. f is going to infinity since sin x >=-1 so f(x) >= x-1 which goes to infinity. g(x) is clearly going to infinity. So we can apply L'Hospital's rule. f'(x)=1+cos x and g'(x)=2x. So lim x ->oo (x+sin x)/x^2 = lim x ->oo (1+cos x)/2x and this is clearly going to 0 since the top stays between 0 and 2 while the bottom goes to infinity.

Now a fun one:
lim x ->oo (x^2+ sin x)/(x^2). I claim we can evaluate this limit without using L'hospital's rule. We have x^2- 1 <= (x^2 + sin x) <= x^2 + 1. Thus, (x^2- 1)/x^2 <= (x^2 + sin x)/x^2 <= (x^2 + 1)/x^2. Finally, since lim x ->oo (x^2- 1)/x^2 = lim x ->oo (x^2 +1)/x^2 = 1 (this is a good exercise if you haven't done any limit problems recently) we have lim x ->oo (x^2+ sin x)/(x^2) =1.

Now, let's see what happens when we apply L'Hospital's rule: Set f(x)= x^2+sin x and g(x)=x^2. Both are going to infinity. f'(x)= 2x+ cos x and g'(x)= 2x. So lim x ->oo (x^2+ sin x)/(x^2) = lim x ->oo (2x+ cos x)/(2x). Ok, this is still infinity over infinity so we can apply L'Hospital's rule again. Thus lim x ->oo (x^2+ sin x)/(x^2) = lim x ->oo (2x+ cos x)/(2x) = lim x ->oo (2 - sin x)/(2), and this limit doesn't exist because sin just keeps oscillating. But the limit has to be 1 by our earlier work. We thus conclude that 1 does not exist.


Elissa said...

Are you sure there isn't a clause in L'Hopital's rule that says "You can take lim(f'/g') unless that limit does not exist"?

Joshua said...

That is essentially the problem. If you go through the proof of L'Hospital's rule it becomes more apparent. One thing that is interesting is that some lower quality high-school textbooks don't even mention this issue in their statements of the theorem.