AP Calculus AB and BC

Limit of ln x / (x - 1) as x Approaches 1

Q: What is the limit of ln x / (x - 1) as x approaches 1?

It is 1, the derivative of \ln x at x = 1.

Q: Is ln x approximately x - 1 near 1?

Yes, that is the tangent line at x = 1. Because \ln x is concave down, x - 1 overestimates it everywhere except at x = 1 itself.

Q: How does this relate to ln(1 + x) / x?

It is the same limit shifted. Substituting x = 1 + u turns one into the other, and both equal 1.

The limit of ln x over x minus 1, as x approaches 1, is 1. The form is 0 over 0, and the expression is the definition of the derivative of ln x at x equals 1. Since that derivative is 1 over x, its value at 1 is 1.

\lim_{x \to 1} \frac{\ln x}{x - 1} = 1

Settled by recognising it as a derivative at a point.

A derivative at x = 1

Since $\ln 1 = 0$ , the numerator is $\ln x - \ln 1$ , and the whole quotient is the alternate form of the derivative of $\ln x$ at $1$ .

\lim_{x \to 1}\frac{\ln x - \ln 1}{x - 1} = \left.\frac{d}{dx}\ln x\right|_{x=1} = \frac{1}{1} = 1

L'Hopital's rule gives the same line of algebra, which is unsurprising: the rule is built out of derivatives.

\lim_{x \to 1}\frac{\ln x}{x-1} \;\overset{\text{H}}{=}\; \lim_{x \to 1}\frac{\frac{1}{x}}{1} = 1

What it says about the graph

The limit is stating that $\ln x$ crosses the $x$ -axis at $x = 1$ with slope exactly $1$ , so its tangent line there is $y = x - 1$ .

That tangent line is the source of a useful approximation: for $x$ near $1$ , $\ln x \approx x - 1$ . Substituting $x = 1 + u$ turns it into the more familiar $\ln(1+u) \approx u$ for small $u$ .

Concavity fixes the direction

The second derivative of ln x is negative one over x squared, so the graph is concave down and the tangent line lies above it. That makes x minus 1 an OVERestimate of ln x for every x other than 1.

The mistakes students make

Substituting $x = 1$ and reporting the limit does not exist. $\frac{0}{0}$ means more work is needed, not that the answer is undefined.
Splitting $\ln x$ over $x - 1$ using a log rule. There is no rule for a logarithm divided by something.
Confusing this with $\lim_{x \to 0}\frac{\ln(1+x)}{x}$ , which is the same fact after a shift, but with the approach at $0$ rather than $1$ .

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of ln x / (x - 1) as x approaches 1?

It is $1$ , the derivative of $\ln x$ at $x = 1$ .

Is ln x approximately x - 1 near 1?

Yes, that is the tangent line at $x = 1$ . Because $\ln x$ is concave down, $x - 1$ overestimates it everywhere except at $x = 1$ itself.

How does this relate to ln(1 + x) / x?

It is the same limit shifted. Substituting $x = 1 + u$ turns one into the other, and both equal $1$ .