limx→e+

(lnx)^

1/

(x−e)

asked by guest
on Oct 19, 2024 at 7:21 am



You asked:

Evaluate the expression: \(\lim_{x \to e^+}{\frac{{\ln\left( x \right)}^{1}}{x - e}}\)

MathBot Answer:

\[\lim_{x \to e^+}{\frac{{\ln\left( x \right)}^{1}}{x - e}} = \infty\]

\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).