## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to \infty} (e^x-cx) = \infty$ $\lim\limits_{x \to -\infty} (e^x-cx) = \infty$ 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.
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.