AP Calculus AB and BC

Limit of ln(x+1)/ln(x) at Infinity Is 1

Q: What about ln(x^2)/ln(x)?

That is exactly 2 for all x > 1, because the exponent comes out as a multiplier.

The limit of ln of x plus 1, over ln x, as x approaches infinity, is 1. One pass of L'Hopital gives x over x plus 1, which tends to 1. Adding a constant inside a logarithm makes no difference to its behaviour at infinity.

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

Settled by L'Hopital's rule.

One pass finishes it

\lim_{x \to \infty}\frac{\ln(x+1)}{\ln x} \;\overset{\text{H}}{=}\; \lim_{x \to \infty}\frac{\frac{1}{x+1}}{\frac{1}{x}} = \lim_{x \to \infty}\frac{x}{x+1} = 1

The algebraic route works too: $\ln(x+1) = \ln x + \ln\left(1+\frac{1}{x}\right)$ , and the second piece tends to $0$ while $\ln x$ grows, so the ratio tends to $1$ .

Logarithms flatten differences

A logarithm compresses everything, so shifts and even multiplicative constants become negligible: $\frac{\ln(kx)}{\ln x} \to 1$ for any positive $k$ as well. Only a change of EXPONENT survives, since $\frac{\ln\left(x^{2}\right)}{\ln x} = 2$ exactly.

The mistakes students make

Cancelling the logarithms to get $\frac{x+1}{x}$ . There is no such rule.
Assuming a shift inside a logarithm matters at infinity. It does not; an exponent does.

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+1)/ln(x) at infinity?

It is $1$ .

What about ln(x^2)/ln(x)?

That is exactly $2$ for all $x > 1$ , because the exponent comes out as a multiplier.