Question-162827

Question Number 162827 by saboorhalimi last updated on 01/Jan/22

Answered by Ar Brandon last updated on 01/Jan/22

g (x) = \lim_{r \to 0} {({(x + 1)}^{r + 1} - x^{r + 1})}^{\frac{1}{r}}

\lim_{x \to \infty} \frac{g (x)}{x} = \lim_{r \to 0, x \to \infty} x^{1 + \frac{1}{r} - 1} {({(1 + \frac{1}{x})}^{r + 1} - 1)}^{\frac{1}{r}}

= \lim_{r \to 0, x \to \infty} x^{\frac{1}{r}} {(1 + \frac{r + 1}{x} - 1)}^{\frac{1}{r}} = \lim_{r \to 0} {(r + 1)}^{\frac{1}{r}}

= \lim_{r \to 0} e^{\frac{1}{r} \ln (r + 1)} = \lim_{r \to 0} e^{\frac{1}{r} (r)} = e