$$ { 解：令x=\frac{1}{n}，则n=\frac{1}{x}；n\rightarrow \infty 时，x\rightarrow 0；于是：} \\ { \lim_{n\rightarrow \infty} \left( n\tan \frac{1}{n} \right) ^{n^2}\begin{array}{l} =\lim_{x\rightarrow 0} \left( \frac{1}{x}\tan x \right) ^{\frac{1}{x^2}}=\lim_{x\rightarrow 0} \left( \frac{\tan x}{x} \right) ^{\frac{1}{x^2}}=\exp \lim_{x\rightarrow 0} \frac{1}{x^2}\cdot \ln \left( \frac{\tan x}{x} \right) =\exp \lim_{x\rightarrow 0} \frac{1}{x^2}\cdot \ln \left[ 1+\left( \frac{\tan x}{x}-1 \right) \right]\\ =\exp \lim_{x\rightarrow 0} \frac{1}{x^2}\cdot \ln \left[ 1+\left( \frac{\tan x}{x}-1 \right) \right] =\exp \lim_{x\rightarrow 0} \frac{1}{x^2}\cdot \left( \frac{\tan x}{x}-1 \right) =\exp \lim_{x\rightarrow 0} \frac{1}{x^2}\cdot \left( \frac{\tan x-x}{x} \right)\\ =\exp \lim_{x\rightarrow 0} \frac{\tan x-x}{x^3}=\exp \lim_{x\rightarrow 0} \frac{\frac{1}{3}x^3}{x^3}=\exp \left( \frac{1}{3} \right) =\mathrm{e}^{\frac{1}{3}}.\\ \end{array}} \\ { 由海涅定理可得：\lim_{n\rightarrow \infty} \left( n\tan \frac{1}{n} \right) ^{n^2}=\mathrm{e}^{\frac{1}{3}}.} \\ { 注：海涅\left( \mathrm{Heine} \right) 定理：设函数f在x_0的一个去心邻域{U}\left( x_0,\delta \right) 中有定义，则} \\ { f在x_0的极限为A，当且仅当对去心邻域{U}\left( x_0,\delta \right) 中任何收敛于x_0的数列\left\{ x_n \right\} ，} \\ { 均有\lim_{n\rightarrow \infty} f\left( x_n \right) =A.} \\ { 因此我们可以认为海涅\left( \mathrm{Heine} \right) 定理构筑了数列极限与函数极限之间的桥梁.} $$