# Use L'Hospital Rule to evaluate the following limits. lim_{xrightarrow0}frac{tanh^{-1}x}{tan(pi x/2)}

Use LHospital
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Given limit function is $\underset{x\to 0}{lim}\frac{{\mathrm{tanh}}^{-1}x}{\mathrm{tan}\left(\pi x/2\right)}$
$\underset{x\to 0}{lim}\frac{{\mathrm{tanh}}^{-1}x}{\mathrm{tan}\left(\pi x/2\right)}=\frac{{\mathrm{tanh}}^{-1}\left(0\right)}{\mathrm{tan}\left(0\right)}=\frac{0}{0}$
Apply LHopitals Rule.
$\underset{x\to 0}{lim}\frac{{\mathrm{tanh}}^{-1}x}{\mathrm{tan}\left(\pi x/2\right)}=\underset{x\to 0}{lim}\frac{2}{\pi {\mathrm{sec}}^{2}\left(\frac{\pi x}{2}\right)\left(1-{x}^{2}\right)}$
$=\frac{2}{\pi }$