The discussion centers on the limit rule involving the natural logarithm and exponential functions as x approaches infinity. It establishes that lim (x->∞) [ln(x^(1/x))] equals 0, and lim (x->∞) x^(1/x) equals 1. Consequently, it confirms that lim (x->∞) [ln(x^(1/x))] can be expressed as ln(lim(x->∞) [(x^(1/x))]), which simplifies to ln(1), resulting in 0. This illustrates the application of limit properties in calculus.
PREREQUISITESStudents of calculus, mathematics educators, and anyone seeking to deepen their understanding of limit rules and their applications in mathematical analysis.
Just what is the question?Rayquesto said:Homework Statement
If lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1, then does this
=>
lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))]) = ln(1)??
Homework Equations
lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1
The Attempt at a Solution
lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))] = ln(1)