The Power of a Reciprocal

As $x$ becomes a very large number ( $x \rightarrow \infty$ ), its reciprocal becomes a very small number ( $\frac{1}{x} \rightarrow 0$ ).

Because $\displaystyle \lim_{x \rightarrow 0 } \alpha^x = 1$ , no matter how big $\alpha$ is, we can state thus that $\displaystyle \lim_{x \rightarrow \infty} x^{\frac{1}{x}} = x^0 = \boxed{1.}$

Oh boy this is a fun one that I learned a while back.

So we can start off by realizing that, by direct substitution of $x = \infty$ , we get an indeterminate $\infty^0$ , so what we need to do is get it into a workable form of $\frac{f(x)}{g(x)}$ so that we can apply l'Hospital's rule, which states...

$\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}$

where both $f(x)$ and $g(x)$ are differentiable on an open interval containing $c$ .

The process by which I manipulate the original function and get it into the proper form is given by the following steps:

$\lim_{x \to \infty} x^{\frac{1}{x}} = \lim_{x \to \infty} e^{\ln(x^{\frac{1}{x}})} = \lim_{x \to \infty} e^{\frac{1}{x}\ln(x)} = \lim_{x \to \infty} e^{\frac{\ln(x)}{x}}$

$\lim_{x \to \infty} e^{\frac{\ln(x)}{x}} = e^{\lim_{x \to \infty}\frac{\ln(x)}{x}}$

Now all we have to do is apply l'Hospital's rule to the $\frac{ln(x)}{x}$ part and we'll have our answer.

$\lim_{x \to \infty}\frac{\ln(x)}{x} = \lim_{x \to \infty}\frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x)} = \lim_{x \to \infty}\frac{\frac{1}{x}}{1} = \lim_{x \to \infty}\frac{1}{x} = 0$

We can then slap that back onto the $e$ superscript to get:

$e^0 = \boxed{1}$