Nice L’Hopital’s Rule Problem

Find the \underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{x}{\cos (x-1)} \right)}^{\frac{x}{1-{{x}^{3}}}}}? 😀


\underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{x}{\cos (x-1)} \right)}^{\frac{x}{1-{{x}^{3}}}}}=\underset{x\to 1}{\mathop{\lim }}\,{{e}^{\frac{x}{1-{{x}^{3}}}\ln \frac{x}{\cos (x-1)}}}=\underset{x\to 1}{\mathop{\lim }}\,{{e}^{x\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1-{{x}^{3}}}\ln \frac{x}{\cos (x-1)}}}

Use The L’Hopital’s Rule for \underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1-{{x}^{3}}}\ln \frac{x}{\cos (x-1)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{\left( \frac{x}{\cos (x-1)} \right)}^{'}}}{{{(1-{{x}^{3}})}^{'}}}=\underset{x\to 1}{\mathop{\lim }}\,\left( \frac{\cos (x-1)+x\sin (x-1)}{-3{{x}^{3}}\cos (x-1)} \right)

Plugging x=1 then we get \underset{x\to 1}{\mathop{\lim }}\,\left( \frac{\cos (x-1)+x\sin (x-1)}{-3{{x}^{3}}\cos (x-1)} \right)=-\frac{1}{3}

Therefore \underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{x}{\cos (x-1)} \right)}^{\frac{x}{1-{{x}^{3}}}}}={{e}^{-\frac{1}{3}}}

Thank you 😀


About Bunchhieng

I like math, programming and web design!!
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