Consider the given expression,

\(\lim_{x\rightarrow0}(\tanh x)^x\)

Since the function does not have any denominator hence, the L'Hospital rule cannot be applied.

Thus,

\(\lim_{x\rightarrow0^+}(\tanh x)^x=\lim_{x\rightarrow0^+}(\frac{e^x-e^{-x}}{e^x+e^{-x}})^x\)

\(=(\frac{e^0-e^{-0}}{e^0+e^{-0}})^0\)

\(=1\)

Which is the required value.

\(\lim_{x\rightarrow0}(\tanh x)^x\)

Since the function does not have any denominator hence, the L'Hospital rule cannot be applied.

Thus,

\(\lim_{x\rightarrow0^+}(\tanh x)^x=\lim_{x\rightarrow0^+}(\frac{e^x-e^{-x}}{e^x+e^{-x}})^x\)

\(=(\frac{e^0-e^{-0}}{e^0+e^{-0}})^0\)

\(=1\)

Which is the required value.