f(x)=ln(x)→∞ as x→∞ (ln(x) grows without bound as x grows without bound) and f(x)=ln(x)→−∞ as x→0+ (ln(x) 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(x)=ln(x) has no horizontal asymptote as x→∞.
Let M>0 be any given positive number (no matter how big). If x>eM, then f(x)=ln(x)>ln(eM)=M (since f(x)=ln(x) is an increasing function). This proves that any horizontal line y=M cannot be a horizontal asymptote of f(x)=ln(x) as x→∞. The fact that f(x)=ln(x) is an increasing function now implies that f(x)=ln(x)→∞ as x→∞.
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(x)=ln(x)<ln(e−M)=−M (since f(x)=ln(x) is increasing). This proves that f(x)=ln(x) gets below any horizontal line if 0<x is sufficiently close to zero. That means f(x)=ln(x)→−∞ as x→0+ .
