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

1 Answer
Apr 9, 2015

f(x)=ln(x)-> \infty as x->\infty (ln(x) grows without bound as x grows without bound) and f(x)=ln(x)->-infty 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->\infty.

Let M>0 be any given positive number (no matter how big). If x>e^{M}, then f(x)=ln(x)>ln(e^{M})=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->\infty. The fact that f(x)=\ln(x) is an increasing function now implies that f(x)=\ln(x)->\infty as x->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(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)->-infty as x -> 0^{+} .