## 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.