x^{1/x} as x approaches infinity? I know that limit of (1+1/x)^{x} = e (as x-> infty), and limit of (1+x)^{1/x} = e (as x-> 0), but this is slightly different.
x^{1/x} as x approaches infinity? I know that limit of (1+1/x)^{x} = e (as x-> infty), and limit of (1+x)^{1/x} = e (as x-> 0), but this is slightly different.
Any help is appreciated
Note that $\displaystyle x^{1/x} = e^{\frac{\ln(x)}{x}}$ so I suggest you apply l'Hospital's theorem to $\displaystyle \frac{\ln(x)}{x}$.
Note that $\displaystyle x^{1/x} = e^{\frac{\ln(x)}{x}}$ so I suggest you apply l'Hospital's theorem to $\displaystyle \frac{\ln(x)}{x}$.
And also note that due to the continuity of the exponential function, that $\displaystyle \displaystyle \lim_{x \to a}e^{f(x)} = e^{\lim_{x \to a}f(x)}$