f(x)=ln(x)-> \inftyf(x)=ln(x)→∞ as x->\inftyx→∞ (ln(x)ln(x) grows without bound as xx grows without bound) and f(x)=ln(x)->-inftyf(x)=ln(x)→−∞ as x -> 0^{+}x→0+ (ln(x)ln(x) grows without bound in the negative direction as xx approaches zero from the right).
To prove the first fact, you essentially need to show that the increasing function f(x)=\ln(x)f(x)=ln(x) has no horizontal asymptote as x->\inftyx→∞.
Let M>0M>0 be any given positive number (no matter how big). If x>e^{M}x>eM, then f(x)=ln(x)>ln(e^{M})=Mf(x)=ln(x)>ln(eM)=M (since f(x)=ln(x)f(x)=ln(x) is an increasing function). This proves that any horizontal line y=My=M cannot be a horizontal asymptote of f(x)=\ln(x)f(x)=ln(x) as x->\inftyx→∞. The fact that f(x)=\ln(x)f(x)=ln(x) is an increasing function now implies that f(x)=\ln(x)->\inftyf(x)=ln(x)→∞ as x->inftyx→∞.
To prove the second fact, let M>0M>0 be any given positive number so that -M<0−M<0 is any given negative number (no matter how far from zero). If 0 < x < e^{-M}0<x<e−M, then f(x)=ln(x)<\ln(e^{-M})=-Mf(x)=ln(x)<ln(e−M)=−M (since f(x)=ln(x)f(x)=ln(x) is increasing). This proves that f(x)=\ln(x)f(x)=ln(x) gets below any horizontal line if 0 < x0<x is sufficiently close to zero. That means f(x)=ln(x)->-inftyf(x)=ln(x)→−∞ as x -> 0^{+}x→0+ .
