AP Calculus AB and BC

Limit of ln(x+1)/x as x Approaches 0 Is 1

Q: Is this the same as the limit of \frac{\ln x}{x-1} as x \to 1?

Yes, after a shift. Put u = x - 1, so x \to 1 becomes u \to 0 and the quotient turns into \frac{\ln(u+1)}{u}, which is this page. Both equal 1, and both are the statement that \frac{d}{dx}[\ln x] = \frac{1}{x} has the value 1 at x = 1.

Q: What is \lim_{x \to 0} \frac{\ln(1+kx)}{x}?

It is k. Multiply and divide by k to rebuild the standard shape: \frac{\ln(1+kx)}{x} = k \cdot \frac{\ln(1+kx)}{kx}, and the second factor tends to 1 as kx \to 0. So \lim_{x \to 0}\frac{\ln(1+5x)}{x} = 5.

Q: Can I just use the approximation \ln(1+x) \approx x?

It gives the right answer, since \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots and dividing by x leaves 1 plus terms that vanish. It is a fine shortcut once the series is available in Unit 10. In Unit 1 it is not, and the approximation is proved from this limit, so the difference quotient is the honest justification.

The limit of ln(x+1)/x as x approaches 0 is 1. Direct substitution gives 0/0, an indeterminate form, so the quotient has to be recognized rather than computed: it is the difference quotient for f(t) = ln(t+1) based at 0, and that derivative is 1/(0+1) = 1.

\lim_{x \to 0} \frac{\ln(x+1)}{x} = 1

Settled by the limit definition of the derivative.

Reading the quotient as a difference quotient

Set $f(t) = \ln(t+1)$ . Then $f(0) = \ln 1 = 0$ , so the numerator is already $f(x) - f(0)$ , and the whole fraction is the difference quotient for $f$ based at the point $0$ with step size $x$ .

\frac{\ln(x+1)}{x} = \frac{f(0+x)-f(0)}{x}, \qquad f(t) = \ln(t+1)

Shrinking the step to $0$ turns a difference quotient into a derivative at the base point, so the limit is $f'(0)$ . The chain rule gives $f'(t) = \frac{1}{t+1}$ , and at $t = 0$ that is $1$ .

\lim_{x \to 0} \frac{\ln(x+1)}{x} = f'(0) = \frac{1}{0+1} = 1

The same value falls out of the definition of $e$ . Since $(1+x)^{1/x} \to e$ as $x \to 0$ , taking a logarithm of that expression produces exactly this quotient.

\ln\left[(1+x)^{1/x}\right] = \frac{\ln(1+x)}{x} \to \ln e = 1

What direct substitution gives

At $x = 0$ the numerator is $\ln(0+1) = \ln 1 = 0$ , and the denominator is $0$ on its own.

\frac{\ln(0+1)}{0} = \frac{\ln 1}{0} = \frac{0}{0}

The form $\frac{0}{0}$ is indeterminate. It reports only that both parts are collapsing, so the value depends on which one collapses faster. The usual algebraic repairs are unavailable here: $\ln(x+1)$ has no polynomial factor of $x$ to pull out, and there is no radical for a conjugate to clear.

The tangent line is the reason

Near $x = 0$ the graph of $y = \ln(x+1)$ hugs the line $y = x$ , so numerator and denominator shrink at matching speed and their ratio settles at $1$ . At $x = 0.001$ the quotient is $0.99950$ , and at $x = -0.001$ it is $1.00050$ .

L'Hopital's rule, and the domain to watch

Since the form is $\frac{0}{0}$ , differentiating the top and the bottom separately is legal and quick.

\lim_{x \to 0} \frac{\ln(x+1)}{x} = \lim_{x \to 0} \frac{\frac{1}{x+1}}{1} = 1

One caution about where the function lives. The logarithm needs $x + 1 > 0$ , so the domain is $x > -1$ , which is a full two-sided neighbourhood of $0$ . The two-sided limit is therefore a fair question, and both sides do return $1$ .

The mistakes students make

Splitting the logarithm as $\ln x + \ln 1$ or as $\ln x + 1$ . A logarithm splits a product, never a sum, so $\ln(x+1)$ stays whole.
Answering $0$ because the numerator goes to $0$ . The denominator goes to $0$ just as fast, and the ratio is what the limit measures.
Cancelling the $x$ inside $\ln(x+1)$ against the $x$ underneath. Nothing cancels across a logarithm.
Reporting that the limit does not exist because the function has no value at $x = 0$ . A limit is decided by nearby inputs only.

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

Is this the same as the limit of $\frac{\ln x}{x-1}$ as $x \to 1$ ?

Yes, after a shift. Put $u = x - 1$ , so $x \to 1$ becomes $u \to 0$ and the quotient turns into $\frac{\ln(u+1)}{u}$ , which is this page. Both equal $1$ , and both are the statement that $\frac{d}{dx}[\ln x] = \frac{1}{x}$ has the value $1$ at $x = 1$ .

What is $\lim_{x \to 0} \frac{\ln(1+kx)}{x}$ ?

It is $k$ . Multiply and divide by $k$ to rebuild the standard shape: $\frac{\ln(1+kx)}{x} = k \cdot \frac{\ln(1+kx)}{kx}$ , and the second factor tends to $1$ as $kx \to 0$ . So $\lim_{x \to 0}\frac{\ln(1+5x)}{x} = 5$ .

Can I just use the approximation $\ln(1+x) \approx x$ ?

It gives the right answer, since $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ and dividing by $x$ leaves $1$ plus terms that vanish. It is a fine shortcut once the series is available in Unit 10. In Unit 1 it is not, and the approximation is proved from this limit, so the difference quotient is the honest justification.