# What is the end behavior of the function f(x) = ln x?

$f \left(x\right) = \ln \left(x\right) \to \setminus \infty$ as $x \to \setminus \infty$ ($\ln \left(x\right)$ grows without bound as $x$ grows without bound) and $f \left(x\right) = \ln \left(x\right) \to - \infty$ as $x \to {0}^{+}$ ($\ln \left(x\right)$ grows without bound in the negative direction as $x$ approaches zero from the right).
To prove the first fact, you essentially need to show that the increasing function $f \left(x\right) = \setminus \ln \left(x\right)$ has no horizontal asymptote as $x \to \setminus \infty$.
Let $M > 0$ be any given positive number (no matter how big). If $x > {e}^{M}$, then $f \left(x\right) = \ln \left(x\right) > \ln \left({e}^{M}\right) = M$ (since $f \left(x\right) = \ln \left(x\right)$ is an increasing function). This proves that any horizontal line $y = M$ cannot be a horizontal asymptote of $f \left(x\right) = \setminus \ln \left(x\right)$ as $x \to \setminus \infty$. The fact that $f \left(x\right) = \setminus \ln \left(x\right)$ is an increasing function now implies that $f \left(x\right) = \setminus \ln \left(x\right) \to \setminus \infty$ as $x \to \infty$.
To prove the second fact, let $M > 0$ be any given positive number so that $- M < 0$ is any given negative number (no matter how far from zero). If $0 < x < {e}^{- M}$, then $f \left(x\right) = \ln \left(x\right) < \setminus \ln \left({e}^{- M}\right) = - M$ (since $f \left(x\right) = \ln \left(x\right)$ is increasing). This proves that $f \left(x\right) = \setminus \ln \left(x\right)$ gets below any horizontal line if $0 < x$ is sufficiently close to zero. That means $f \left(x\right) = \ln \left(x\right) \to - \infty$ as $x \to {0}^{+}$ .