Answer
$\lim\limits_{x \to \infty} (e^x-cx) = \infty$
$\lim\limits_{x \to -\infty} (e^x-cx) = \infty$ if $c \gt 0$
$\lim\limits_{x \to -\infty} (e^x-cx) = 0$ if $c = 0$
$\lim\limits_{x \to -\infty} (e^x-cx) = -\infty$ if $c \lt 0$
$f(x)$ has an absolute minimum if $c \gt 0$
As $c$ increases, the minimum points decrease as they become more negative.
Work Step by Step
We can find the limit as $x \to \infty$:
$\lim\limits_{x \to \infty} (e^x-cx) = \lim\limits_{x \to \infty} x(\frac{e^x}{x}-c)$
Using L'Hospital's Rule:
$\lim\limits_{x \to \infty} \frac{e^x}{x} = \lim\limits_{x \to \infty} \frac{e^x}{1} = \infty$
Therefore:
$\lim\limits_{x \to \infty} x(\frac{e^x}{x}-c) = \infty$
We can find the limit as $x \to -\infty$:
$\lim\limits_{x \to -\infty} (e^x-cx) = \lim\limits_{x \to -\infty} x(\frac{e^x}{x}-c)$
Note that $\lim\limits_{x \to -\infty} \frac{e^x}{x} = 0$
Therefore:
$\lim\limits_{x \to -\infty} x(\frac{e^x}{x}-c) = \lim\limits_{x \to -\infty} x(0-c) = \lim\limits_{x \to -\infty} -cx$
This limit goes to $\infty$ if $c \gt 0$
This limit goes to $0$ if $c = 0$
This limit goes to $-\infty$ if $c \lt 0$
$\lim\limits_{x \to -\infty} (e^x-cx) = \infty$ if $c \gt 0$
$\lim\limits_{x \to -\infty} (e^x-cx) = 0$ if $c = 0$
$\lim\limits_{x \to -\infty} (e^x-cx) = -\infty$ if $c \lt 0$
We can find the value of $x$ such that $f'(x) = 0$:
$f'(x) = e^x-c = 0$
$e^x = c$
$x = ln(c)$
$f[ln(c)] = e^{ln~c}-c~(ln~c) = c~(1-ln~c)$
As $c$ increases, the minimum points decrease as they become more negative.
Note that $(ln~c)$ is defined if $c \gt 0$
When $c \gt 0$:
$\lim\limits_{x \to \infty} (e^x-cx) = \infty$
$\lim\limits_{x \to -\infty} (e^x-cx) = \infty$
Therefore, $f(x)$ has an absolute minimum if $c \gt 0$
As $c$ increases, the minimum points decrease as they become more negative.
