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

Question

Accepted Answer

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&#x27;Hopital&#x27;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}})

Answer

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&#x27;Hopital&#x27;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}})

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

Answer

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