# How do you use L'hospital's rule to find the limit lim_(x->oo)x^(1/x) ?

The basic idea in using the rule of De l'Hospital to find indeterminate limits of powers $f {\left(x\right)}^{g \left(x\right)}$ is to rewrite it as ${e}^{g \left(x\right) \setminus \ln \left(f \left(x\right)\right)}$ and find the limit of the indeterminate product $g \left(x\right) \ln \left(f \left(x\right)\right)$ rewriting the product as a quotient: $\ln \frac{f \left(x\right)}{\frac{1}{g} \left(x\right)}$ or g(x)/(1/(ln(f(x)))
If the power was indeterminate (${0}^{0}$ or ${1}^{\infty}$ or ${\infty}^{0}$) then the obtained quotient is either indeterminate of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, so that the Rule of De l'Hospital applies to lift the indetermination.
In this example ${x}^{\frac{1}{x}} = {e}^{\frac{1}{x} \ln x}$ and ${\lim}_{x \setminus \to \infty} \ln \frac{x}{x} = {\lim}_{x \to \infty} \frac{\frac{1}{x}}{1} = 0$ by the Rule of de l'Hospital.
Thus${\lim}_{x \setminus \to \setminus \infty} {x}^{\frac{1}{x}} = {e}^{0} = 1$