Consider the limit \(f\left( x \right) = {e^x} - cx\).
Differentiate the function w.r.t \(x\).
\(\begin{array}{c}f'\left( x \right) = \frac{d}{{dx}}\left( {{e^x} - cx} \right)\\ = {e^x} - c\end{array}\)
As \(f'\left( x \right) = 0\) then we have,
\(\begin{array}{c}0 = {e^x} - c\\{e^x} = c\\x = \ln c\end{array}\)
where \(c > 0\). Now differentiate \(f'\left( x \right) = {e^x} - c\) then we have, \(f''\left( x \right) = {e^x}\).
Since \(f''\left( x \right) = {e^x} > 0\). So \(f\) is concave up on the interval \(\left( { - \infty ,\infty } \right)\).
Now find the limit.
\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {{e^x} - cx} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{d}{{dx}}\left( {{e^x} - cx} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {x\left( {\frac{{{e^x}}}{x} - c} \right)} \right)\\ = {L_1}\end{array}\)
Now apply l’Hospital’s Rule.
\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{c} = \mathop {\lim }\limits_{x \to \infty } \frac{d}{{dx}}\left( {\frac{{{e^x}}}{c}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^x}}}{c}} \right)\\ = \infty \end{array}\)
Therefore \({L_1} = \infty \).
\(L = \mathop {\lim }\limits_{x \to - \infty } \left( {{e^x} - cx} \right),\;{e^x} \to 0\)thus \(L\) is determined by \( - cx\).
If \(c > 0\), \( - cx \to \infty \) and \(L = - \infty \).
Thus \(f\) has an absolute minimum for \(c > 0\). As the value of \(c\) increases, the minimum points \(\left( {\ln c,c - c\ln c} \right)\) get farther away from the origin.
