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Infinite limit of ln(x) as x approaches 0 from the right

Consider the natural logarithm function f(x)=ln⁡(x)f(x) = \ln(x).

Examine the behavior of f(x)=ln⁡(x)f(x) = \ln(x) as x→0+x \to 0^+.

Infinite limit of ln(x) as x approaches 0 from the right graph

Evaluate lim⁡x→0+ln⁡(x)\displaystyle\lim_{x \to 0^+} \ln(x), using +∞+\infty or −∞-\infty notation if appropriate. Then state whether x=0x = 0 is a vertical asymptote of f(x)=ln⁡(x)f(x) = \ln(x), and justify your conclusion in one sentence.

Hint

Think about what happens to ln⁡(x)\ln(x) as xx takes on smaller and smaller positive values, like 0.1,0.01,0.0010.1, 0.01, 0.001. Recall that ln⁡(x)\ln(x) is only defined for x>0x > 0, so only the right-hand limit makes sense here.

Step-by-step solution

As xx approaches 00 from the right, we plug in progressively smaller positive values such as x=0.1,0.01,0.001x = 0.1, 0.01, 0.001, which give ln⁡(0.1)≈−2.303\ln(0.1) \approx -2.303, ln⁡(0.01)≈−4.605\ln(0.01) \approx -4.605, and ln⁡(0.001)≈−6.908\ln(0.001) \approx -6.908. These values are decreasing without bound, showing that ln⁡(x)\ln(x) grows more and more negative as x→0+x \to 0^+.

This unbounded negative growth means the limit does not settle on a finite number, so we write lim⁡x→0+ln⁡(x)=−∞\displaystyle\lim_{x \to 0^+} \ln(x) = -\infty. Since this one-sided limit is infinite and f(x)=ln⁡(x)f(x) = \ln(x) is undefined at x=0x = 0, the vertical line x=0x = 0 is a vertical asymptote of the graph.

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