Right-hand infinite limit and vertical asymptote

As $x \to 1^+$, we have $x$ slightly larger than $1$, so $\ln(x)$ is a small positive number approaching $0$ from the positive side, i.e. $\ln(x) \to 0^+$. Taking the reciprocal of a small positive number produces a very large positive number, so $1/\ln(x)$ grows without bound.

Therefore $\lim_{x \to 1^+} \frac{1}{\ln(x)} = +\infty$. Since this one-sided limit is infinite and $f(x) = 1/\ln(x)$ is undefined at $x=1$ (because $\ln(1) = 0$), the graph of $f$ has a vertical asymptote at $x = 1$.

As x approaches one from the left, x is slightly less than one, so the natural log of x is a small negative number approaching zero from below. Taking the reciprocal of a small negative number gives a large negative number. Therefore the function decreases without bound, approaching negative infinity from the left side of x equals one.