The limit is (famously) [tex]e[/tex] by definition, but if you want to compute it via L'Hopital's rule:
[tex]\displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x=\lim_{x\to\infty}\exp\left(\ln\left(1+\frac1x\right)^x\right)=\exp\left(\lim_{x\to\infty}x\ln\left(1+\frac 1x\right)\right)=\exp\left(\lim_{x\to\infty}\frac{\ln\left(1+\frac1x\right)}{\frac1x}\right)[/tex]
where [tex]\exp(x)=e^x[/tex]. As [tex]x\to\infty[/tex], the numerator approaches [tex]\ln1=0[/tex] and the denominator approaches 0. Applying L'Hopital's rule gives
[tex]\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)[/tex]
The remaining approaches 1, so the original limit is [tex]e^1=e[/tex], as expected.
