We all know that $e$, by definition, is $\lim_{x\to \infty}(1+\frac{1}{x})^x$.
But $\lim_{x\to \infty}(a^\frac{1}{x} -\frac{1}{x}) =1$ for all a $$\\ \implies \lim_{x\to \infty}a^\frac{1}{x}=\lim_{x\to \infty }(1+\frac{1}{x})\\ \implies a=\lim_{x\to \infty} (1+\frac{1}{x})^x={e}$$
If this is correct, then $e$ can be any number, which is definitely not true. Tell me which step is wrong.
The part where you say that $$ \lim_{x\to \infty} a^{1/x} = \lim_{x\to\infty}(1+1/x) \implies a = \lim_{x\to\infty}(1+1/x)^x.$$
Taking each side to the power $x$ gives $$\left(\lim_{x\to \infty} a^{1/x}\right)^x = \left(\lim_{x\to\infty}(1+1/x) \right)^x$$
There is no justification for exchanging the order of the limit and $y\mapsto y^{x},$ and this 'proof of nonsense' is an excellent example of why. (Also note that the $x$ in the limit is a "dummy variable", so it arguably doesn't even make sense to take both sides to the $x$-th power... what is this "$x$" we're talking about?)
We need to evaluate the limit inside first, giving $1^x=1^x,$ which is of course true. The fact that moving the $x$ inside gives you something different is just proof that that move is illegal.
