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}}}}}? 😀

Solution

\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 😀

Advertisements

About Bunchhieng

I like math, programming and web design!!
This entry was posted in Problems. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s