$\lim _{x \rightarrow \infty}(1-\tanh x)^{\frac{1}{x}}$

Step 1 ：Rewrite the expression as: \(\lim _{x \rightarrow \infty}e^{\frac{1}{x}\ln(1-\tanh x)}\)

Step 2 ：Find the limit of the exponent: \(\lim _{x \rightarrow \infty}\frac{\ln(1-\tanh x)}{x}\)

Step 3 ：Apply L'Hopital's rule: \(\lim _{x \rightarrow \infty}\frac{\ln(1-\tanh x)}{x} = \lim _{x \rightarrow \infty}\frac{-\text{sech}^2 x}{1-\tanh x}\)

Step 4 ：Calculate the limit of the exponent: \(-2\)

Step 5 ：Find the limit of the original expression: \(\lim _{x \rightarrow \infty}(1-\tanh x)^{\frac{1}{x}} = e^{\lim _{x \rightarrow \infty}\frac{\ln(1-\tanh x)}{x}} = e^{-2}\)

Step 6 ：\(\boxed{e^{-2}}\)