Elementary proof of the prime number theorem? The prime number theorem is equivalent to $\lim_{x \to \infty} \dfrac{1}{x} \left| \sum_{n\leq x} \mu(n) \right| = 0$, where $\mu(n)$ is the Mobius function. 
We know that $\left| \sum_{n\leq x} \mu(n) \right| \leq x$ for all integers $x\geq 1$, and  since this grows in a certainly nonlinear way, we can write $\left| \sum_{n\leq x} \mu(n) \right| \leq x^{1-\delta}$ for every $x\geq 2$. For the purposes of our argument, it suffices to assume that $0<\delta<1$ is dependent on $x$.. Dividing throughout by $x$ we arrive at                                                         $\dfrac{1}{x}\left| \sum_{n\leq x} \mu(n) \right| \leq x^{-\delta}$. Taking limits as $x$ tends to $\infty$, we obtain the desired result ?
 A: The proof starts to go off the rails here:

...and since this grows in a certainly nonlinear way, we can write $\left|\sum_{n\leq x}\mu(n)\right|\leq x^{1-\delta}$ for every $x\geq 2$.

Okay, that's true - though it only encompasses the fact that $\left|\sum_{n\leq x}\mu(n)\right|<x$, rather than capture any idea that that function grows "in a nonlinear way" (which is asserted without proof or even definition).  Moreover, you really ought to be writing something like:
$$\left|\sum_{n\leq x}\mu(n)\right|\leq x^{1-\delta(x)}$$
since $\delta$ is dependent on $x$. But then dividing by $x$ gives
$$\frac{1}x\left|\sum_{n\leq x}\mu(n)\right|\leq x^{-\delta(x)}$$
and taking limits as $x$ goes to infinity gives
$$\lim_{x\rightarrow\infty}\frac{1}x\left|\sum_{n\leq x}\mu(n)\right|\leq \lim_{x\rightarrow\infty}x^{-\delta(x)}$$
but exactly what the right hand side is depends on the function $\delta$. For instance, if $\delta(x)$ equaled $\frac{1}{\log(x)}$ then the left hand converges to $\frac{1}{e}$, which is decidedly positive, contrasting with your proof, where you take this limit to be $0$.
Moreover, one might note that this proof has used nothing about the prime numbers - if I replaced the Mobius function with a function $\mu(n)$ defined to equal $1$ when $n$ is even and $0$ otherwise, then nothing in your proof breaks, but it claims that $\lim_{x\rightarrow\infty}\frac{1}x\left|\sum_{n\leq x}\mu(n)\right|=0$
which is obviously false.
