Let $f(x) =c \log(x) -x$ for some $c>0$. Find $\lim_{x\to \infty} f(x) $
Intuitively it seems the required limit is $-\infty$, I am trying to make it precise. It is clear that $f'(x) $ approaches to $-1$ for large values of $x$ but i cannot see how to use this.
We can use the formula $\lim\limits_{x\to\infty}\frac{\log(x)}{x} = 0\\$ (e.g., apply L' Hopital)
Note that $f(x) = x\cdot \left(\frac{c\log(x)}{x}-1\right)$ and therefore the limit can be written as \begin{align*} \lim\limits_{x\to\infty}f(x) &= \lim\limits_{x\to\infty}\left(x\cdot \left(\frac{c\log(x)}{x}-1\right)\right)\\ &= \lim\limits_{x\to\infty}x\cdot \lim\limits_{x\to\infty}\left(\frac{c\log(x)}{x}-1\right)\\ &= \lim\limits_{x\to\infty}x\cdot \left(c\left(\lim\limits_{x\to\infty}\frac{\log(x)}{x}\right)-1\right)\\ &= \lim\limits_{x\to\infty}x\cdot(0 - 1)\\ &= \lim\limits_{x\to\infty}(-x)\\ &= -\infty \end{align*}
Another variant:
$g(x) = e^{f(x)} = {e^{\log x^c} \over e^x}$
$\lim_{x \rightarrow \infty} g(x) = {e^{\log x^c} c/x\over e^x} = 0$ (from L'Hopital's rule, since it is an $\infty/\infty$ form)
From here you can conclude that $\lim_{x \rightarrow \infty} f(x) = -\infty$.
