Proof that lim x-> inf (x ^ (1/x)) = 1

I started this proof by letting f(x) = (x ^ (1/x)) - 1 and attempting to show that that converges to zero. It's pretty straightforward to establish that a limit exists--x > e^1 => f(x) > 0 and f'(x) < 0, so f is bounded below and decreasing. Now what? Should I try lim inf, knowing that the existence of a finite limit implies that that limit equals the lim inf?

Re: Proof that lim x-> inf (x ^ (1/x)) = 1

Quote:

Originally Posted by

**phys251** I started this proof by letting f(x) = (x ^ (1/x)) - 1 and attempting to show that that converges to zero. It's pretty straightforward to establish that a limit exists--x > e^1 => f(x) > 0 and f'(x) < 0, so f is bounded below and decreasing. Now what? Should I try lim inf, knowing that the existence of a finite limit implies that that limit equals the lim inf?

Notation: $\displaystyle \exp(t)=e^t$.

So $\displaystyle {\lim _{x \to \infty }}{x^{\frac{1}{x}}} = {\lim _{x \to \infty }}\exp \left( {\frac{{\ln (x)}}{x}} \right) = \exp \left( {{{\lim }_{x \to \infty }}\frac{{\ln (x)}}{x}} \right) = \exp \left( 0 \right) = 1$

Re: Proof that lim x-> inf (x ^ (1/x)) = 1

Oh wow. Take e^(ln(f(x)), and get the limit inside the exp(), triggering L'Hopital's rule. Thanks!