## Calculus (3rd Edition)

$\displaystyle\lim_{r\rightarrow 0} (1+r)^{1/r}=e$
We have to estimate the limit: $\displaystyle\lim_{r\rightarrow 0} (1+r)^{1/r}$ Compute $(1+r)^{1/r}$ for values of $r$ close to $0$, approaching from both sides: $(1+(-0.01))^{1/(-0.01)}\approx 2.731999$ $(1+(-0.001))^{1/(-0.001)}\approx 2.7196422$ $(1+(-0.0001))^{1/(-0.0001)}\approx 2.7184178$ $(1+(-0.00001))^{1/(-0.00001)}\approx 2.7182954$ $(1+(-0.000001))^{1/(-0.000001)}\approx 2.7182832$ $(1+0.000001)^{1/0.000001}\approx 2.7182805$ $(1+0.00001)^{1/0.00001}\approx 2.7182682$ $(1+0.0001)^{1/0.0001}\approx 2.7181459$ $(1+0.001)^{1/0.001}\approx 2.7169239$ $(1+0.01)^{1/0.01}\approx 2.7048138$ Therefore we got: $\displaystyle\lim_{r\rightarrow 0} (1+r)^{1/r}=e$