Function exercise asked in OMEC as entrance exam in 2015 -2016

Exercise:

Let $f:\mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R}$ defined to be $f\left( x \right)=x-\ln \left| x \right|$

1) Study the limit of $f$ on the boundary of domain and deduce its asymptotes

2) Determine the derivative of this function and its variation table

3) Trace the curve of $f$

4) Calculate, $\int_{1}^{e}{f\left( x \right)dx}$

Solution:

1) $\underset{x\to -\infty }{\mathop{\lim }}\,f\left( x \right)=\underset{x\to -\infty }{\mathop{\lim }}\,x-\ln \left| x \right|=-\infty -\infty =-\infty $

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x-\ln \left| x \right|=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x-\ln \left( -x \right)={{0}^{-}}-\ln \left( -{{0}^{-}} \right)=\infty $

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x-\ln \left| x \right|$ as $x\to {{0}^{+}}\Leftrightarrow \left| x \right|=x$

So $=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x-\ln x={{0}^{+}}-\ln {{0}^{+}}=0+\infty =\infty $ thus $x=0$ is V.A

$\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,x-\ln \left| x \right|=\infty -\infty $ ind form

$=\underset{x\to \infty }{\mathop{\lim }}\,x\left( 1-\frac{\ln \left| x \right|}{x} \right)=\infty \left( 1-\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left| x \right|}{x} \right)=\infty $  as $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left| x \right|}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln x}{x}=0$

Let $y=\ln x\Leftrightarrow x={{e}^{y}}$ as $x\to \infty ,y\to \infty $ So $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln x}{x}=\underset{y\to \infty }{\mathop{\lim }}\,\frac{y}{{{e}^{y}}}$

We know that ${{e}^{y}}\ge \frac{{{y}^{2}}}{2}\Leftrightarrow 0<\frac{y}{{{e}^{y}}}\le \frac{2}{y}\Leftrightarrow \underset{y\to \infty }{\mathop{\lim }}\,\frac{y}{{{e}^{y}}}=0$

2) $f'\left( x \right)=1-\frac{d}{dx}\ln \left| x \right|$ but $\left| x \right|=\sqrt{{{x}^{2}}}$

So $\frac{d}{dx}\ln \left| x \right|=\frac{d}{dx}\ln \left( \sqrt{{{x}^{2}}} \right)=\frac{\frac{d}{dx}\left( \sqrt{{{x}^{2}}} \right)}{\sqrt{{{x}^{2}}}}=\frac{\frac{2x}{2\sqrt{{{x}^{2}}}}}{\sqrt{{{x}^{2}}}}=\frac{x}{\sqrt{{{x}^{4}}}}=\frac{x}{{{x}^{2}}}=\frac{1}{x}$

Thus $f'\left( x \right)=1-\frac{1}{x}$ so $f'\left( x \right)=0\Leftrightarrow \frac{1}{x}=1\Leftrightarrow x=1$

We have $f'\left( x \right)=\frac{x-1}{x}$

\(\begin{align}
  & \underline{\left. \,\,\,\,\,\,\,x\,\,\, \right|-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\
 & \underline{\left. x-1\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,} \\
 & \underline{\left. \,\,\,\,\,x\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,} \\
 & \underline{\left. f'\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,} \\
 & \left. f\,\,\,\,\,\,\,\,\,\, \right|\,{{\,}_{-\infty }}\,\,\,\nearrow \,{{\,}^{+\infty }}\left. {} \right\|\,{{\,}^{+\infty }}\,\,\,\searrow \,\,{{\,}_{_{_{\,f\left( 1 \right)=1}}}}\,\,\,\nearrow \,{{\,}^{+\infty }} \\
\end{align}\)

4- ) $\int_{1}^{e}{f\left( x \right)dx}=\int_{1}^{e}{\left( x-\ln \left| x \right| \right)dx}=\int_{1}^{e}{xdx}-\int_{1}^{e}{\ln \left| x \right|dx}$

Since $1\le x\le e\Leftrightarrow \left| x \right|=x$

So $\int_{1}^{e}{\ln \left| x \right|dx}=\int_{1}^{e}{\ln x}\,dx=\left[ x\ln x-x \right]_{1}^{e}=\left( e-e-\left( 1\times 0-1 \right) \right)=0+1=1$

And $\int_{1}^{e}{xdx}=\left[ \frac{{{x}^{2}}}{2} \right]_{1}^{e}=\frac{{{e}^{2}}}{2}-\frac{1}{2}$

Thus $\int_{1}^{e}{f\left( x \right)dx}=\frac{{{e}^{2}}}{2}-\frac{1}{2}-1=\frac{{{e}^{2}}}{2}-\frac{3}{2}=\frac{{{e}^{2}}-3}{2}$