Answer
$\displaystyle\lim_{r\rightarrow 0} (1+r)^{1/r}=e$
Work Step by Step
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:
$\displaystyle\lim_{r\rightarrow 0} (1+r)^{1/r}=e$
