Question

Limx-> (1+(1/x))^× using L'hospital​

Answer 1

The limit is (famously)
`e` by definition, but if you want to compute it via L'Hopital's rule:

where
`\exp(x)=e^x`. As
`x\to\infty`, the numerator approaches
`\ln1=0` and the denominator approaches 0. Applying L'Hopital's rule gives

`\displaystyle\exp\left(\lim_(x\to\infty)\frac{-\frac1{x^2}\cdot\frac1{1+\frac1x}}{-\frac1{x^2}}\right)=\exp\left(\lim_(x\to\infty)\frac1{1+\frac1x}\right)`

The remaining approaches 1, so the original limit is
`e^1=e`, as expected.